text:
Last Update on 22 August, 2019
The Children's Online Privacy Protection Act of 1998 and its rules (collectively, "COPPA") require us to inform parents and legal guardians (as used in this section, "parents") about our practices for collecting, using, and disclosing personal information from children under the age of 13 ("children"). It also requires us to obtain verifiable consent from a child's parent for certain collection, use, and disclosure of the child's personal information.
This section notifies parents of:
The types of information we may collect from children.
Our practices for disclosing that information.
Our practices for notifying and obtaining parents' consent when we collect personal information from children, including how a parent may revoke consent.
All operators that collect or maintain information from children through this website.
This section only applies to children under the age of 13 and supplements the other provisions of this privacy policy. Only the other provisions of this privacy policy apply to teens and adults.
Terms that are defined in the general privacy policy have the same meanings as used in this Privacy Policy for Children Under the Age of 13.
Information We Collect from Children
Children can access many parts of the website and the Learning Platform and its content and use many of its features without providing us with personal information. However, some content and features are available only to registered users or require us to collect certain information, including personal information, from them. In addition, we use certain technologies, such as cookies, to automatically collect information from our users (including children) when they visit or use the website and the Learning Platform.
We only collect as much information about a child as is reasonably necessary for the child to participate in an activity, and we do not condition his or her participation on the disclosure of more personal information than is reasonably necessary.
Information We Collect Directly. A child must provide us with the following information to register with this website: the child's user name and password and a parent's email address. We also require the child to create a member name and password. We may request additional information from your child, but this information is optional. We specify whether information is required or optional when we request it.
Automatic Information Collection and Tracking. We use technology to automatically collect information from our users, including children, when they access and navigate through the website and the Learning Platform and use certain of its features. The information we collect through these technologies may include:
One or more persistent identifiers that can be used to recognize a user over time and across different websites and online services.
We also may combine non-personal information we collect through these technologies with personal information about you or your child that we collect online.
For information about our automatic information collection practices, including how you can opt out of certain information collection, see the "Information We Collect" and "How You Can Control Your Privacy" sections of this Privacy Policy.
How We Use Your Child's Information
We use the personal information we collect from your child to:
register him or her with the website and the Learning Platform;
communicate with him or her about activities or features of the website and the Learning Platform that may be of interest;
We use the information we collect automatically through technology (see "Information We Collect") and other non-personal information we collect to improve our website and the Learning Platform and to deliver a better and more personalized experience by enabling us to:
Store information about the child's preferences, allowing us to customize the content according to individual interests.
Our Practices for Disclosing Children's Information
We do not share, sell, rent, or transfer children's personal information other than as described in this section.
We may disclose aggregated information about many of our users and information that does not identify any individual or device. In addition, we may disclose children's personal information:
To third parties we use to support the internal operations of our website and the Learning Platform and who are bound by contractual or other obligations to use the information only for such purpose and to keep the information confidential.
If we are required to do so by law or legal process, such as to comply with any court order or subpoena or to respond to any government or regulatory request.
If we believe disclosure is necessary or appropriate to protect the rights, property, or safety of WiselyWise, our customers or others, including to:
protect the safety of a child;
protect the safety and security of the website and the Learning Platform; or
enable us to take precautions against liability.
To law enforcement agencies or for an investigation related to public safety.
Accessing and Correcting Your Child's Personal Information
At any time, you may review the child's personal information maintained by us, require us to correct or delete the personal information, and/or refuse to permit us from further collecting or using the child's information.
You can review, change, or delete your child's personal information by:
Logging into your child's account and visiting his or her account profile page.
Sending us an email at chandru@wiselywise.com. To protect your privacy and security, we may require you to take certain steps or provide additional information to verify your identity before we provide any information or make corrections.
Any other inquiries may be directed to:
WiselyWise Pte. Ltd.
28 Upper Serangoon View #15-33
Email: chandru@wiselywise.com
This Privacy Statement may be updated periodically and will be posted on our website indicating when it was last updated. If there are material changes to our Privacy Statement, we will post prominent notice on our website and/or provide other notice as required by law.
You can contact us by email or at the postal address below. Please indicate in your email or letter that you are making a Privacy or Privacy Statement Inquiry.


text:
Virtual Workarounds Help African Couples Marry Across Distance
web desk September 7, 2020 September 8, 2020
JOHANNESBURG – Marriage. In these socially distanced times, even the ceremony itself can't always bring couples — and their families — together. In Africa, some digitally savvy couples are finding virtual workarounds to get them to the altar, including weddings where the bride and groom are thousands of kilometers apart.
Fifi and Emmanuel Egbebu got married last month.
But in these difficult times of coronavirus restrictions on travel, they resorted to a modern solution: they did it virtually.
She was in Nigeria's southeastern Enugu state, dancing herself down the aisle for their traditional Nigerian wedding.
And he was in Johannesburg, South Africa's economic hub, where he works as a musician, watching it all online.
The bride says her special day was no less special despite the absence of the groom.
"On that day, he was, like online, and we're talking every second," she said. "So, I didn't feel like he wasn't around. I actually felt he was with me, like, he was with me, really, just that he wasn't there, present, you know. I'm sure people will be wondering, 'Why is this girl dancing and laughing and having fun?' I was just I just felt he was really there. It wasn't any different to me. I didn't feel different. I felt good."
Egbebu notes that a close friend of hers did the same recently, virtually marrying with the groom in England and the bride in Nigeria.
They're among the countless couples across the world kept apart by the pandemic.
But even those who are together can't celebrate in a traditional way because of pandemic restrictions on large crowds.
It's crashed the business of Johannesburg party supplier Molebogeng Gulwako.
She supplies weddings and engagement parties with "champagne walls," transparent shelves of filled glasses in front of an ornate wall.
Gulwako, herself a newlywed, says it may take some cultures on this side of the world a while to accept virtual celebrations.
"It's going to take a very long time. Because remember, we also grew up in big families," Gulwako said. "We always had people — at least a minimum of seven people — around you for your seventh, eighth 15th and 25th birthday and so forth. So, it's going to be very hard to unlearn, celebrating on your own. I'm not sure about the Western culture, because maybe you can just do a dinner with one other person and that's OK. But that's not us. That's not me."
While some couples are delaying their weddings until COVID-19 restrictions are lifted, the Egbebus didn't want to wait to declare their love and commitment.
"I felt bad that I couldn't make it there in person. At the same time, when I saw her dance, I was smiling. It kind of gave me confidence I needed, that she's happy," said Emmanuel Egbebu. "So at least that's something to hold on to."
The Egbebus plan to reunite in South Africa when flights resume.
For now, these two — and countless others — will have to have and hold each other virtually.
Source: Voice of America
Woman defrauded of R1.7 million pension money, suspect in court
RevBits annonce l'émission d'un brevet américain pour une architecture unique de produit de sécurité utilisant le cryptage zéro connaissance

redpajama_set_name: RedPajamaCommonCrawl


text:
Oracle's Business Intelligence Suite Reborn?
The revamped Business Intelligence Suite incorporates analytic technology from the former Siebel Systems Inc., which Oracle acquired 16 months ago.
It was just about a year ago that Oracle Corp. unveiled Oracle Business Intelligence (Business Intelligence) Suite, a revamped, enlarged, and Project Fusion-ized version of its erstwhile Oracle Business Intelligence offering.
At the time, some industry watchers groused that Oracle Business Intelligence Suite promised a Business Intelligencet more than it actually delivered. There may have been something to that, but with Monday's release of a revamped Business Intelligence Suite offering, Oracle hopes to put such grumblings to rest.
The retooled package, dubbed Oracle Business Intelligence Suite Enterprise Edition 10g R3, incorporates analytic technology from the former Siebel Systems Inc., which Oracle acquired 16 months ago. The new R3 release amounts to an incremental upgrade over Siebel 7.8, even as it adds significantly to the functionality of the Business Intelligence Suite Enterprise Edition which Oracle first introduced last March.
Nor does the new Business Intelligence Suite incorporate the ETL and data integration assets of the former Sunopsis Inc., which Oracle nabbed late last year. Not yet, at least: earlier this month, Oracle unveiled a re-badged version of the former Sunopsis technology, dubbed Oracle Data Integrator (ODI), on its Oracle Technology Network Web site. Unlike Oracle Warehouse Builder (OWB), ODI is designed to complement Oracle's Project Fusion middleware layer.
Oracle plans to provide more details about ODI in the coming weeks, officials promise. In the meantime, they say there's plenty for new and existing customers to explore in the retooled Oracle Business Intelligence Suite.
Oracle Does Right By Its Namesake
Legend has it that the priestesses at Delphi and at other ancient oracles used to prophesy instantaneously, with little or no pause for reflection or cognition.
Call it real-time decision-making, circa the first millennium BCE.
Business Intelligence-driven decision-making isn't quite as turnkey. And in this respect, Oracle's Business Intelligence technology—like that of its competitors—has charted a more torturous course to real-time nirvana. The reason, experts say, is that real-time Business Intelligence—or, more precisely, the loosely coupled data and application infrastructure that (in most cases) empowers real-time decision-making—is a lot more complicated than spontaneous divination.
That's why Project Fusion gives Oracle a demonstrable edge, officials claim. True, it might have been born of necessity—i.e., as a means to integrate Oracle's disparate PeopleSoft, J.D. Edwards, and classic applications—but Project Fusion also gave Oracle an opportunity to optimize the integration between and among its constituent applications.
The result, experts say, is a cohesive Business Intelligence Suite whose whole is much greater than the sum of its disparate parts. "Unlike products such as Oracle Business Intelligence Discoverer, Oracle Reports Services, and Oracle Portal, the query and delivery tools within Business Intelligence Suite Enterprise Edition were built from the ground up to share the same metadata, application server platform, calculation infrastructure and administration tools," writes Oracle consultant Mark Rittman.
Nor does R3 disappoint on the real-time tip. Nearly a year into its Project Fusion make-over, the revamped Oracle Business Intelligence Suite delivers what Oracle officials claim are bona-fide real-time capa Business Intelligencelities—including a real-time Interactive Dashboard component. "We made some enhancements to the Oracle Interactive Dashboards, which make those dashboards even more interactive [so that users can] change presentation variables on-the-fly, change the value in real-time and have the analytics be calculated on-the-fly," says Rick Schultz, vice-president of Fusion Middleware with Oracle. "They [are] instantaneous [calculations]. Before, there just wasn't an easy way to quickly sub in a new value and have the entire set of information recalculated based on that new variable. Now there is."
Other improvements include support for conditional chart formatting, PDF generation (such that users can generate PDF files from snapshots of their interactive dashboards), and JSR168 portlet support. Oracle Answers—the Business Intelligence Suite's ad hoc query and analysis tool—also gets an R3-ready revamping, says Schultz: "We've added a drag-and-drop layout editor [and] we've added a calculations editor to make it easier for end-users to modify a formula."
While a lot of new Oracle Business Intelligence Suite features came from Siebel, its revamped reporting component—which boasts pixel-perfect reporting capaBusiness Intelligencelities—is homegrown, officials claim. "The reporting came from Oracle, [what used to be called] Oracle XML Publisher, which we now call Oracle Business Intelligence Publisher. It supports pixel-perfect analysis and reporting," Schultz indicates. "So you can do documents to forms to checks, statements, invoices, labels, etc. It also supports editing and production in Word and Excel and Adobe Acrobat."
Speaking of Word, Excel, and other Office productivity tools, the revamped Business Intelligence Suite Enterprise Edition boasts improved Office integration, too. New features include support for most Office output formats, enhanced data export capaBusiness Intelligencelities, and live interaction with Word, Excel, and Powerpoint.
Elsewhere, Oracle officials touted features like improved data mart automation—such that users can create data models directly from business views and automatically generate the physical data mart structure—improved performance (including native 64-Business Intelligencet support, a new hypercube infrastructure, and multi-pass calculation optimizations and others), and other niceties.
Business Intelligenceg Deal or Business Intelligenceg Brouhaha?
So just how Business Intelligenceg of a deal is Oracle Business Intelligence Suite Enterprise Edition R3 10g? Oracle consultant Rittman says it's all a matter of perspective.
"This is more or less what Siebel themselves planned to release as part of the Siebel 8.0 stack, but it's got Oracle's XML Publisher integrated—[although it's] now known as Business Intelligence Publisher—plus it's got some integration points with Oracle Fusion Middleware," Rittman says. "It's not dramatically different [than] Siebel Analytics 7.8; it's more of an update to the existing product rather than a complete overhaul—this will come with the 11g release in 2008."
In this respect, Rittman says, while R3 is a significant improvement over the first rev of Oracle's Business Intelligence Suite, it's only an incremental improvement, at best, over Siebel's next planned analytic release. "[R3] is a Business Intelligenceg improvement over the previous offering," if by "previous offering" one means Oracle Discoverer, Oracle Reports, and other legacy offerings, he says. "If … you mean Siebel Analytics 7.8, then it's not a major overhaul—this will come with 11g."


This refined 16th century palace on the Grand Canal offers amazing views and has one of the largest gardens at the back. An ideal venue for a wedding reception or gala dinner.
Guest Numbers: Up to 120 guests.
Garden: One of the loveliest gardens in Venice.
Music: Loud music is prohibited in the garden. Live music and DJs are welcome until 1am on the ground floor.
Accommodation: One apartment accommodates 6 people and the other 4.


text:
Insurers need to reinvent in terms of agency, service
Ever since the President gave his nod to the ordinance on the Insurance Bill, the industry has been euphoric.
Written by guest
Updated: February 3, 2015 1:35:47 pm
Ever since the President gave his nod to the ordinance on the Insurance Bill, the industry has been euphoric. The sector, which grew rapidly after nationalisation and, more so, after it was opened to the private sector in 2000, is now in turmoil due to unabated mis-selling.
Premiums have become the metric to assess performance. This is evident from LIC's focus on single premium, which accounts for 60% of the behemoth's business. For the private sector, single premium accounts for around 30% of the total premium. Most private insurers have replicated LIC's distribution model, going with the same products. The chunk of the business continues to be garnered by agents, who have been quitting thick and fast.
While the regulator has come out strongly to protect consumer interest, it now needs to focus on the security of lives. Insurance requires simplicity in evolution and distribution of products. To stay ahead of the curve, the industry will have to reinvent itself in terms of agency and servicing.
It needs to relook at its priorities in bringing security for people at an affordable cost.
The high cost of distribution calls for a total revamp of the agency profession, instead of a blind selection of candidates. The industry need to refine the selection process. As it is difficult to train people, selections based on a scientific approach can prove to be more productive. The industry needs to identify those with strong interpersonal skills and the passion to serve people.
High attrition
The high attrition of agency makes it highly expensive. Unfortunately, the industry has not found it expedient to record the costs involved in its books, notwithstanding the fact that profits can be indeterminate without assessing the impact of the cost of each input. The access to business analytics should have made the task easy.
Agency regulations need to be reoriented to make selling easy and sustainable. There should be an internship at the college level to assess its acceptability. The candidates identified can be allowed apprenticeship to sell simple products based on risk-like term insurance. The volumes should take care of earnings. After all, Public Provident Fund was a runaway success on even 1% commission to agents.
After training and working for a few months, agents can graduate to endowment products and unit-linked insurance products (Ulips). In fact, Ulips have been exploited by both agents and companies, transfer as they do risk to consumers. It is prudent to sell Ulips only to those who have some exposure to the stock market.
Agents need to act as financial advisors and they should acquire the capacity to deliver. Financial literacy campaigns, awards and sponsorship of events in educational institutions will create a favourable impact. Substantial micro insurance needs have been taken care of by the Jan Dhan Yojana. However, the payment of insurance premium can be allowed as an expense under CSR for marketing micro-insurance to identified groups by suitable amendments.
Empowering customers through technology
Several services are available at the click of a button. The Interactive Voice Response system of insurers will give details on policy status, due date, loan and surrender value. As a result, intervention by agents can be minimised. E-governance should enable outsourcing such activities to generate employment, and the Digital India campaign can give a fillip. Mobile apps can be developed to boost efficiency. The industry has been lacking in training modules on websites.
By RR Grover
The author is director, Amity School of Insurance, Banking & Actuarial Science and former chief of marketing, LIC
Insurance Bill
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Gold price rises as investors flock to safe-haven metals, but the window for up-move may close soon


Parenting is one of the most challenging, demanding jobs there is. It is can be one of the most rewarding – though it doesn't always feel like that. We believe that ideas from this course can make the job easier, more fun and more survivable when the going gets tough.
Come along and find out how.
The course runs for 10 weeks from Monday 10 September 2018 from 2pm 'til 3pm.
It will be held at Alver Valley Schools.


text:
int vel = 1;
int acc = 2;
int ret = 0;
double joint1;

wrock::cube_msg cube_cmd;

void jointCallback(const sensor_msgs::JointState::ConstPtr& msg)
{
  joint1 = msg->position[4];

}

int main(int argc, char** argv)
{
  ros::init(argc, argv, "cube_demo");

  ros::NodeHandle n;
    ros::Publisher cmd_pub = n.advertise<wrock::cube_msg>("joint_cmd", 100);
ros::Publisher pub = n.advertise<sensor_msgs::PointCloud2> ("assembled_cloud", 1);
    ros::Subscriber joint_sub = n.subscribe("joint_states", 1, jointCallback);
    // Create the service client for calling the assembler
  ros::ServiceClient    client = n.serviceClient<laser_assembler::AssembleScans2>("assemble_scans2");
laser_assembler::AssembleScans2 srv;
joint1 = 1;
cube_cmd.j1 = 2.5;
cube_cmd.j2 = 0;
cube_cmd.j3 = 0;
cube_cmd.j4 = 0;
cube_cmd.j5 = 0;
cube_cmd.vel = 1.5;
cube_cmd.acc = 2;
cube_cmd.pose = true;
        //send the joint state and transform
    // Populate our service request based on our timer callback times
cmd_pub.publish(cube_cmd);
ros::spinOnce();
printf("j1: %f\n",joint1);
ROS_INFO("going to start");
while(joint1 < 2.4){
ros::spinOnce();
//printf("j1: %f\n",joint1);
}
 srv.request.begin = ros::Time::now();
cube_cmd.j1 = 0;

cube_cmd.vel = 0.25;
cmd_pub.publish(cube_cmd);
ros::spinOnce();
ROS_INFO("waiting for end");
while(joint1 > 0){
ros::spinOnce();
}
ROS_INFO("done");
 srv.request.end   = ros::Time::now();
    // Make the service call
    if (client.call(srv))
    {
      ROS_INFO("Published Cloud ", (uint32_t)(srv.response.cloud.width)) ;
      pub.publish(srv.response.cloud);
    }
    else
    {
      ROS_ERROR("Error making service call\n") ;
    }
ros::spinOnce();
while(ros::ok()){
}
  return 0;
}

redpajama_set_name: RedPajamaGithub


text:
Q: SH2016 patch config Wiz error I am trying to implement the new Jun patch on SH2016 farm and when I try to run the config Wiz I get an error unable to upgrade [searchAdminDatabase] failed and the reason it's looking for a user account which was disabled, when I checked the DB owner I noticed all search, profile, Social, state service, sync, Wss_usageapplication DBs are owned by this account, it looks like this account was used to install SHP. All other DB are owned by Farm account.
I tried to change the ownership of these DBs to Farm account in SQL but did not fix the problem. My question is there a way to change the DB owner to continue or I have to install SHP again.

A: You should not disable the service account that was used to install SP, this can cause all kinds of issues.  Also it is typically recommended that you use this installer service account to install CUs.  For quick fix, if this is possible,  I'd enable the service account and try the wizard again.

A: I hope you have changed the farm account from the Central Administrator site as well.

*

*Go to Central admin

*Open Security page and click on Service Account, from the drop down change the farm account to the new account. Run IIS reset and restart SharePoint time job.

redpajama_set_name: RedPajamaStackExchange


Tinsels is a story with four hands, those of Rémi and Julien.
drawing inspiration from music and encounters, express the Tinsels spirit.
The brand moved to Lyon where the culture of beautiful fabric, well-made clothing resonate as an inheritance.


text:
Special assessments—Election—Notes.
(1) The board of directors may, at any time when in their judgment it may be advisable, call a special election and submit to the qualified electors of the district the question whether or not a special assessment shall be levied for the purpose of raising money to be applied to any of the purposes provided in this chapter including any purpose for which the bonds of the district or the proceeds thereof might be lawfully used. Such election must be called upon the notice prescribed, and the same shall be held and the result thereof determined and declared in all respects in conformity with the provisions of RCW 87.03.200. The notice must specify the amount of money proposed to be raised and the purpose for which it is intended to be used and the number of installments in which it is to be paid. At such election the ballot shall contain the words "Assessment Yes" and "Assessment No." If the majority of the votes cast are "Assessment Yes" the board may immediately or at intervals thereafter incur indebtedness to the amount of said special assessment for any of the purposes for which the proceeds of said assessment may be used, and may provide for the payment of said indebtedness by the issue and sale of notes of the district to an amount equal to said authorized indebtedness, which notes shall be payable in such equal installments not exceeding three in number as the board shall direct. Said notes shall be payable by assessments levied at the time of the regular annual levy each year thereafter until fully paid. The amount of the assessments to be levied shall be ascertained by adding fifteen percent for anticipated delinquencies to the whole amount of the indebtedness incurred and interest. Each assessment so levied shall be computed and entered on the assessment roll by the secretary of the board, and collected at the same time and in the same manner as other assessments provided for herein, and when collected shall be paid to the county treasurer of the county to the credit of said district, for the purposes specified in the notice of such special election: PROVIDED, HOWEVER, That the board of directors may at their discretion issue said notes in payment for labor or material, or both, used in connection with the purposes for which such indebtedness was authorized. Notes issued under this section shall bear interest at a rate determined by the board, payable semiannually. Such notes may be in any form, including bearer notes or registered notes as provided in RCW 39.46.030.
(2) Notwithstanding subsection (1) of this section, such notes may be issued and sold in accordance with chapter 39.46 RCW.
[ 1983 c 167 § 220; 1981 c 156 § 28; 1921 c 129 § 24; 1915 c 179 § 19; 1895 c 165 § 24; 1889-90 p 692 § 41; RRS § 7458. Formerly RCW 87.32.110.]
Liberal construction—Severability—1983 c 167: See RCW 39.46.010 and note following.
Assessments: RCW 87.03.240 through 87.03.255, 87.03.265 through 87.03.305.
Ballots in all elections: RCW 87.03.075.
Elections are governed by irrigation district laws: RCW 87.03.030.


Willy Cigars is a cigar company based in Chandler, Arizona. Led by Cuban-born craftsman Guillermo "Willy" Carrazana, Willy Cigars produces a variety of high-quality, handmade cigars. With over 45 years of experience in the cigar trade, Willy has cultivated an ability to expertly create cigars. He personally selects and imports tobacco leaves from the Dominican Republic, and he blends these cigars with Cuban tobacco seeds in order to ensure optimal aroma and taste.
Originally hailing from Cuba, Willy first entered the cigar crafting business in 1958 at the age of eight. He was initially mentored by his grandfather, who taught him the process of hand making high-quality cigars. Later, Willy came to work in a cigar factory in Cuba, where he refined his knowledge of cigar making, eventually immigrating to the United States and settling in Arizona.


Sadly, it seems we just can't get through an entire Major League Baseball season without at least one tragic fan death.
Two seasons ago it was the Rangers fan who fell over a railing while reaching out for a ball thrown by Josh Hamilton. Last year it was a Blue Jays fan who died of a heart attack down in the field boxes. And, unfortunately, this year it is a 30-year-old Atlanta Braves fan named Ronald Lee Homer Jr.
During a rain delay in Monday night's game against the Philadelphia Phillies at Turner Field, Homer somehow went over a railing on the upper level of the stadium and fell 65 feet to the parking lot below. The police were called immediately, but he was later pronounced dead at the hospital.
It's still unknown whether alcohol played any role in his death, but police do believe it was an accident.
If there's any silver lining to this at all, it's that Ronald Homer spoke to his mother not long before he died and told her he loved her.
Of course, Homer's death comes just one week after a Kansas City woman managed to cheat death during an ill-advised swim in the fountains at Kauffman Stadium.
Let's hope this tragedy will serve as a reminder for fans everywhere to be careful.


text:
Q: Expressão Regular para validar senha - PHP Pesquisei aqui no SO e tbm no Google, mas não encontrei algo que agisse como eu gostaria, também não sou bom com expressões regulares, confesso que tenho certa dificuldade e gostaria de fazer uma validação de senha em PHP que:

*

*Tenha pelo menos 8 caracteres (isso eu verifico usando strlen());

*Possua pelo menos 1 número;

*Possua pelo menos 1 letra maiúscula;

*Possua pelo menos 1 letra minúscula;

*Possua pelo menos 1 caractere especial;

Se alguém puder me dar uma luz, ficarei grato.

A: Já tem um bom tempo, mas vai aqui uma possibilidade de validar apenas com um regex puro:
/(?!\s)(?=.*?[a-z]{1,})(?=.*?[A-Z]{1,})(?=.*?\d{1,}).*(?=.?[^a-zA-Z0-9]{1,})?.([^\s]){8,}/g

No último parênteses, dentro do colchete, você pode excluir todos os caracteres que quiser, por exemplo, se quiser remover aspas:
([^\s'"`])

Testando:

Explicando um pouco:
Quando você quer mais de uma regra no regex, onde precise passar por todas, você utiliza (?=seu_regex) (seria similar ao && em linguagens de programação).
Como nesse caso, não importa a ordem do que foi digitado, desde que tenha todos os caracteres, então pra cada regra desejada, você utiliza: .*?, algo como "independente de quantos ou que vem antes, valide minha regra". Logo:
(?=.*?seu_regex)

Quando você precisa quantificar suas regras, é só utilizar {n} para uma quantidade exata, ou {n,} para "a partir de". Então, se quisesse obrigar no mínimo quatro letras minúsculas, {4,}. Seguindo o caso da sua pergunta:
(?=.*?seu_regex{1,})


text:
FP Street
OSC's Sino-Forest secrecy questioned as cease-trade order extended
Why, if the information produced was damning enough to prompt the regulator to act so swiftly and aggressively, was the firm not compelled by the regulator to release it to the investing public?
Barbara Shecter
Sep 08, 2011 • Last Updated September 8, 2011 • 2 minute read
Norm Betts/Bloomberg
Howard Wetston, chair of the Ontario Securities Commission, knows about the importance of disclosure. After all, he chaired the OSC panel that penned a much-studied decision on the importance of full and continuous disclosure in 2003 following a probe at YBM Magnex International Inc.
As well-versed in the subject as Mr. Wetston is — now in his second stint at Canada's largest capital markets watchdog — it came as a surprise to some in Bay Street's tight-knit legal community that he and his agency slapped a cease trade order on the shares of embattled Sino-Forest Corp. last month.
The OSC sought Thursday morning to extend the initial order until at least Jan. 25.
Why, if the information produced was damning enough to prompt the regulator to act so swiftly and aggressively, was the firm not compelled by the regulator to release it to the investing public? As some senior securities lawyer see it, it was out of step with a long-held view at the regulator that its role is to ensure investors have access to as much relevant information as possible on which to base their investment decisions.
For much of the summer, until the moment the OSC ordered shares of Sino to be cease traded on Aug. 26, investors were trading based on contradictory claims and little in the way of fact.
Short-seller Muddy Waters claimed Sino overstated its assets and was operating a fraud. The company countered that it was all a misunderstanding that would be explained when a special committee of independent directors had completed an internal probe.
On Aug. 26, the investing public still did not know what had been unearthed by Sino's independent committee so far. But it appears the OSC did, having received sufficient information to declare that the company appears to have "misrepresented some of its revenue" and "exaggerated some of its timber holdings." The regulator further alleges that some officers and directors engaged in acts they "know or reasonably ought to know perpetuate a fraud."
An affidavit sworn by OSC senior forensic accountant Anthony Long in connection with Thursday's hearing reveals OSC staff were informed on the afternoon of Aug. 24 that two Sino vice-presidents had been temporarily suspended. The next day, OSC staff learned of the suspension of a third vice-president.
Karen Manarin, a lawyer for OSC staff, argued Thursday that the cease-trade order was necessary because further disclosure of information investors would require to make "informed investment decisions" could "seriously compromise" the regulator's ongoing investigation, as well as the probe undertaken by Sino's independent committee.
The chair of the OSC hearing, Mary Condon, approved the extension of the cease trade order, which was not contested by the company.
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I periodically tell my 6 year old that I am her boyfriend.
Gee Angus, you must have met my father! .....and by the way Richard, the answer to that is not much longer!


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\begin{document}

\title[On the improvements of Hardy and Copson inequalities]
{On the improvements of Hardy and Copson inequalities}
\author[Das]{Bikram Das}
\address{Dr. APJ Abdul Kalam Technical University, Lucknow-226021 \&
Indian Institute of Carpet Technology, Chauri Road, Bhadohi- 221401, Uttar Pradesh, India}
\email{dasb23113@gmail.com}
\author[Manna]{Atanu Manna$^*$ }
\address{Indian Institute of Carpet Technology, Bhadohi-221401, Uttar Pradesh, India}
\email{atanu.manna@iict.ac.in, atanuiitkgp86@gmail.com ($^*$Corresponding author)}

\subjclass[2010]{Primary 26D15; Secondary 46A45.}
\keywords{ Discrete Hardy's inequality; Improvement; Copson's inequality; Sequence space.}

\begin{abstract}
In this current work, we revisit the recent improvements of the discrete Hardy's inequality in one dimension and establish an extended improved discrete Hardy's inequality with its optimality. We also study one-dimensional discrete Copson's inequality (E.T. Copson, \emph{Notes on a series of positive terms}, J. London Math. Soc., 2 (1927), 9-12.), and achieve an improvement of the same in a particular case. Further, we study some fundamental structures such as completeness, K\"{o}the-Toeplitz duality, separability, etc. of the sequence spaces which originated from the improved discrete Hardy and Copson inequalities in one dimension.
\end{abstract}
\maketitle

\section{Introduction}{\label{secint}}
The famous discrete Hardy's inequality was developed in the twentieth century during the period $1906-1928$. Apart from the contribution of G. H. Hardy, the other mathematicians such as E. Landau, G. P\'{o}lya, I. Schur, M. Riesz contributed a lot to the development of it. A systematic survey on the prehistory of Hardy's inequality is well-explained by Kufner et al. \cite{AKR}. Let us recall the one-dimensional discrete Hardy's inequality in its crudest form. For $p>1$, and a sequence $\{a_{n}\}_{n=1}^{\infty}$ of complex numbers the classical discrete Hardy's inequality (\cite{GHYLITTLE}, Theorem 326) in one dimension states that
\begin{align}{\label{DHI}}
&\displaystyle\sum_{n=1}^{\infty}\Big|\frac{1}{n}\sum_{k=1}^{n}a_{k}\Big|^{p}<\Big(\frac{p}{p-1}\Big)^{p}\displaystyle\sum_{n=1}^{\infty}|a_{n}|^{p},
\end{align} holds unless $a_n$ is null. Here $p>1$ is a real no., and the constant term $\big(\frac{p}{p-1}\big)^{p}$ associated with the inequality (\ref{DHI}) is sharp. In a letter, Landau (see \cite{ELU}) wrote to G.H. Hardy, and addressed a proof of the inequality (\ref{DHI}) with a sharp constant. The results of Landau were officially published five years later than the letter of Landau to Schur \cite{LANDAU}. During this period Hardy \cite{GHYNOTE} was working with both continuous as well as discrete cases of Hardy's inequality, and later he also commented on Landau's letter. Since both E. Landau and G. H. Hardy have contributed to the development of the inequality (\ref{DHI}), so this inequality (\ref{DHI}) sometimes called as Hardy-Landau inequality \cite{AKR}.\\
The 1925 articles of G. H. Hardy (\cite{GHYNOTE251}, \cite{GHYNOTE252}) contain many interesting results. One of them is the following extension of Hardy's inequality (\ref{DHI}). Suppose that $\{q_n\}$ sequences of real numbers such that $q_{n}>0$, and denote $A_{n}=q_{1}a_{1}+ q_{2}a_{2}+ \ldots +q_{n}a_{n}$ and $Q_{n}=q_1+q_2+\ldots+q_{n}$ for $n\in \mathbb{N}$. If for $p>1$, $\{q_{n}^{\frac{1}{p}}a_{n}\}\in \ell_p$ then
\begin{align}{\label{GDHI}}
\displaystyle\sum_{n=1}^{\infty}q_nQ_{n}^{-p}|A_{n}|^{p}&< \Big(\frac{p}{p-1}\Big)^{p}\displaystyle\sum_{n=1}^{\infty}q_{n}|a_{n}|^{p},
\end{align} unless all $a_n$ is null. Also the constant term $\big(\frac{p}{p-1}\big)^{p}$ is sharp. The inequality (\ref{GDHI}) is studied and extended in various ways. For instance, E. T. Copson \cite{ECN} introduced and studied an extended version of inequality (\ref{GDHI}) as below. Let $1<c\leq p$. Then
\begin{align}{\label{COPI}}
&\displaystyle\sum_{n=1}^{\infty}q_{n}Q^{-c}_{n}|A_{n}|^{p}\leq\Big(\frac{p}{c-1}\Big)^{p}\displaystyle\sum_{n=1}^{\infty}q_{n}Q^{p-c}_{n}|a_{n}|^{p},
\end{align}where the associated constant term is best possible, equality holds good when all $a_n$ are `$0$'. There are many applications of these inequalities (\ref{DHI}), (\ref{GDHI}), and (\ref{COPI}) can be found in several parts of analysis in the form of generalization, extensions or their direct applications to spectral theory, graph theory, differential equations etc.. A careful study on the analysis and geometry of Hardy's inequality can be found in \cite{BEL}. For various studies on the Hardy inequalities, we refer to \cite{MBN}, \cite{GHYLITTLE}, \cite{KAPLAP}, \cite{LEINGEN}, \cite{PLEFEVRE}, \cite{LIUZ}, \cite{MANNANEW}, and references cited therein for the readers. One of the recent trends of research in operator theory is to improve various inequalities for operators on Hilbert space. Many authors are engaging themselves in investigating improvements of the well-known numerical radius and Berezin number inequalities for bounded linear operators on a Hilbert space $\mathcal{H}$ (see \cite{MAJEE} for the latest research). But as far as the discrete Hardy's inequality is concerned, we couldn't find any point-wise improvement
in previous years. In 2018, we have a surprising result on point-wise improvement of the discrete classical Hardy's inequality (\ref{DHI}), and it is due to Keller et al. \cite{MKR}. To write briefly their results, we rewrite the Hardy inequality (\ref{DHI}) in a different form, and use the following notation.\\
Denote $C_{c}(\mathbb{N}_0)$ as the space of finitely supported functions on $\mathbb{N}_0=\{0, 1, 2, 3, \ldots\}$. Then clearly for all $A=(A_n)\in C_{c}(\mathbb{N}_0)$, where $A_n=a_1+a_2+\ldots+a_{n}$ with the assumption that $A_0=0$, inequality (\ref{DHI}) is equivalent to the following:
\begin{align}{\label{EFDHI}}
&\displaystyle\sum_{n=1}^{\infty}|A_n-A_{n-1}|^{p}\geq\Big(\frac{p-1}{p}\Big)^{p}\displaystyle\sum_{n=1}^{\infty}\frac{|A_n|^{p}}{n^{p}}.
\end{align}
Recently, Keller et al. \cite{MKR} improved the inequality (\ref{EFDHI}) for the case when $p=2$, and later Fischer et al. \cite{FFR} obtained the improvement of (\ref{EFDHI}) for general $p>1$. In fact the authors in \cite{MKR} have proved that
\begin{align}{\label{IMHI}}
\displaystyle\sum_{n=1}^{\infty}| A_n-A_{n-1}|^{2}&\geq\frac{1}{4}\displaystyle\sum_{n=1}^{\infty} \frac{| A_n|^{2}}{n^{2}}+\displaystyle\sum_{k=2}^{\infty}\binom{4k}{2k}\frac{1}{(4k-1)2^{4k-1}}\displaystyle\sum_{n=2}^{\infty}\frac{|A_n|^{2}}{n^{2k}}>\frac{1}{4}\displaystyle
\sum_{n=1}^{\infty} \frac{|A_n|^{2}}{n^{2}},
\end{align}and which is equivalent to the following
\begin{align}{\label{IMHI2}}
\displaystyle\sum_{n=1}^{\infty}| A_n-A_{n-1}|^{2}&\geq\displaystyle\sum_{n=1}^{\infty} w_n|A_n|^{2}>\frac{1}{4}\displaystyle
\sum_{n=1}^{\infty} \frac{|A_n|^{2}}{n^{2}},
\end{align}where the sequence $w_n$ is defined as follows:
\begin{center}
$w_n=2-\sqrt{1-\frac{1}{n}}-\sqrt{1+\frac{1}{n}}>\frac{1}{4n^2}$, $n\in \mathbb{N}$.
\end{center}
The history of defining such kind of sequence is also very surprising and pretty interesting. The readers are referred to the works of Keller et al. (\cite{MKR}, \cite{MKRGRAPH}) for getting a history. In a recent work of Krej\v{c}i\v{r}\'{i}k and \v{S}tampach \cite{DKK}, a short and elementary proof of inequality (\ref{IMHI}) is presented by proving an identity, which finally contributes a remainder term. The optimality of the sequence $w_n$ is also established by these authors. In a more latest result of Krej\v{c}i\v{r}\'{i}k et al. (see Theorem 10, \cite{DKALFS}), the authors established that
\begin{align}{\label{IMHIGN3}}
\displaystyle\sum_{n=1}^{\infty} w_n(g)|A_n|^{2}&\leq \displaystyle\sum_{n=1}^{\infty}| A_n-A_{n-1}|^{2},
\end{align}
which is an extension of inequality (\ref{IMHI2}), and the sequence $w_n(g)$ is defined as below
\begin{center}
$w_n(g)=2-\frac{g_{n-1}}{g_n}-\frac{g_{n+1}}{g_n}$, $n\in \mathbb{N}$,
\end{center}where $g_n$ is a strictly positive sequence of real numbers with the convention that $g_0=0$. In particular, when $g_n=\sqrt{n}$ one can easily have the inequality (\ref{IMHI2}). The criteria for optimality of $w_n(g)$ was also considered by these researchers \cite{DKALFS}. Using the idea of defining $w_n(g)$, an improvement of Rellich inequality was obtained by Gerhat et al. \cite{BGDKFS}, which is an unpublished preprint now. \\
Observe that, if we substitute $p=2$, $\alpha=2-c$ and $q_n=1$ for all $n\in \mathbb{N}$ in (\ref{COPI}), then it transformed into the following:
\begin{align*}
\displaystyle\sum_{n=1}^{\infty}{n}^{\alpha-2}|A_{n}|^{2}&\leq\frac{4}{(\alpha-1)^{2}}\displaystyle\sum_{n=1}^{\infty}{n}^{\alpha}|a_{n}|^{2},
\end{align*}which further equivalent to the following inequality:
\begin{align}{\label{COPIPAR}}
\displaystyle\sum_{n=1}^{\infty}|A_{n}-A_{n-1}|^{2}{n}^{\alpha}&\geq \frac{(\alpha-1)^2}{4}\sum_{n=1}^{\infty}\frac{|A_{n}|^{2}}{n^2}{n}^{\alpha},
\end{align}where it is pre-assumed that $A_0=0$. A very surprising and noticeable point is that Gupta \cite{SGA} has recently investigated the improvement of inequality (\ref{COPIPAR}), which is called one-dimensional discrete Hardy's inequality with power weights. Indeed Gupta \cite{SGA} obtained the following inequality:
\begin{align}{\label{GUPI}}
\displaystyle\sum_{n=1}^{\infty}|A_{n}-A_{n-1}|^{2}n^{\alpha} &\geq \displaystyle\sum_{n=1}^{\infty}w_{n}(\alpha,\beta)|A_{n}|^{2},
\end{align}
where $\alpha, \beta\in \mathbb{R}$, $w_{1}(\alpha,\beta):=1+2^{\alpha}-2^{\alpha+\beta}$, and for $ n\geq2$
\begin{align*}
w_{n}(\alpha,\beta)& =n^{\alpha}\Big[1+\Big(1+\frac{1}{n}\Big)^{\alpha}-\Big(1-\frac{1}{n}\Big)^{\beta}-\Big(1+\frac{1}{n}\Big)^{\alpha+\beta}\Big].
\end{align*}
The importance of inequality (\ref{GUPI}) is of two types, one is it contains Hardy's inequality (\ref{COPIPAR}) with power weights whenever $\alpha\in [0, 1)\cup[5,\infty)$, and another is the improvement of (\ref{COPIPAR}) for the case when $\alpha\in [\frac{1}{3}, 1)\cup \{0\}$.
\begin{Remark}
It is to be noted that in the case when $\alpha=0$, and $\beta=\frac{1}{2}$, then inequality (\ref{GUPI}) transformed into the improved discrete Hardy's inequality (\ref{IMHI}) derived by Keller et al. \cite{MKR}.
\end{Remark}
With the above discussions, one may ask the following
\begin{center}
\emph{(a) Is it possible to extend both the inequalities (\ref{IMHIGN3}), and (\ref{GUPI}) for more general weight sequence, say $\{\lambda_n\}$?}
\end{center}
\vspace{0.2cm}
Now let us choose another substitutions $p=2$, $\alpha=2-c$, and $q_{n}=n$ for all $n\in \mathbb{N}$ in inequality (\ref{COPI}), we have
\begin{align*}
\displaystyle \sum_{n=1}^{\infty}S_n^{\alpha}\frac{|A_{n}-A_{n-1}|^{2}}{n} &\geq \frac{(\alpha-1)^{2}}{4}\displaystyle \sum_{n=1}^{\infty}\frac{n}{S_n^{2-\alpha}}|A_{n}|^{2},
\end{align*}or equivalently
\begin{align}{\label{COPIQNEQN}}
\displaystyle \sum_{n=1}^{\infty}S_n^{2-c}\frac{|A_{n}-A_{n-1}|^{2}}{n} &\geq \frac{(c-1)^{2}}{4}\displaystyle \sum_{n=1}^{\infty}\frac{n}{S_n^{c}}|A_{n}|^{2},
\end{align} where $S_n=\frac{n(n+1)}{2}$.\\
We then have two natural questions as below:
\begin{center}
\emph{(b) Is it possible to improve inequality (\ref{COPIQNEQN})?},
\emph{and if the answer is affirmative then} \\
\emph{(c) What are values of $c$ $(1<c\leq 2)$ for which improvement of (\ref{COPIQNEQN}) possible?}
\end{center}
Therefore, the aim of this present note is two folds. In one fold of this paper, we will answer question \emph{(a)} posed above, and establish a generalized improved discrete Hardy's inequality in one dimension. This extension not only includes the inequality (\ref{IMHI}) of Keller et al. \cite{MKR} but also contains the recent inequality (\ref{IMHIGN3}) of Krej\v{c}i\v{r}\'{i}k et al. (see Theorem 10, \cite{DKALFS}) as well as the inequality (\ref{GUPI}) of Gupta \cite{SGA}. In another fold, we will be concentrated on the answers to questions \emph{(b)}, and \emph{(c)}. We prove that improvement of Copson's inequality (\ref{COPIQNEQN}) is possible, and moreover, we can prove that it is true only when $c=\frac{3}{2}$.\\
The paper is organized as follows, in section 2 we prove an extended improved one-dimensional Hardy's inequality, and derive several consequences. In section 3, we will be dealing with Copson's inequality and its possible improvements and optimality of the weight sequence. In addition to these results, section 4, provides some fundamental properties of the sequence space $\Gamma_p$, $p>1$ which is created from the improved Hardy inequality (\ref{IMHI2}), and Copson inequality (\ref{IMCOP}).

\section{Generalization of Improved discrete Hardy's inequality}
We now proceed to present one of the main results of this paper. First, we begin with the following theorem.
\begin{thm}{\label{THMIDHI}}
Let $A_n$ be any sequence of real or complex numbers such that $A_n\in C_c(\mathbb{N}_0)$ with $A_{0}=0$ and $g=\{g_{n}\}_{n=1}^{\infty}$ be any strictly positive sequence of real numbers. Then the following inequality holds:
\begin{align}{\label{GIMHI}}
\displaystyle\sum_{n=1}^{\infty}w_{n}(\lambda,g )|A_n|^{2}\leq\displaystyle\sum_{n=1}^{\infty}\frac{|A_{n}-A_{n-1}|^{2}}{\lambda_{n}},
\end{align} where $\lambda=\{\lambda_n\}_{n\geq 1}$ such that $\lambda_{n}>0$, $n\in \mathbb{N}$, and the sequence $w_{n}(\lambda,g)$ is defined as below:
\begin{align*}
w_{n}(\lambda,g)& =\frac{1}{\lambda_{n}}+\frac{1}{\lambda_{n+1}}-\frac{g_{n-1}}{\lambda_{n}g_{n}}-\frac{g_{n+1}}{\lambda_{n+1}g_{n}}.
\end{align*}
Further, if there exists a sequence of elements $\gamma^{N}\in C_{c}(\mathbb{N}_{0})$ such that $\gamma^{N}\leq\gamma^{N+1}$ with $\gamma^{N}\rightarrow1$ as $N\rightarrow\infty$ pointwise, and
\begin{align}{\label{LIMCOND}}
\displaystyle\lim_{N\rightarrow\infty}\sum_{n=2}^{\infty}\frac{g_{n}g_{n-1}}{\lambda_{n}}|\gamma^{N}_{n}-\gamma^{N}_{n-1}|^{2}=0,
\end{align}
then $w_{n}(\lambda, g)$ is optimal.
\end{thm}
\begin{proof}
To prove the inequality (\ref{GIMHI}), we first choose $h_{n}=g_{n}-g_{n-1}$ so that $h_{n+1}=g_{n+1}-g_{n}$. Then $w_{n}(\lambda, g)$ can be re-written as follows:
\begin{center}
$w_{n}(\lambda,g)=\frac{h_{n}}{\lambda_{n}g_{n}}-\frac{h_{n+1}}{\lambda_{n+1}g_{n}}$.
\end{center}
We observe that
\begin{align*}
\displaystyle\sum_{n=1}^{\infty}w_{n}(\lambda,g)|A_{n}|^{2}
& =\displaystyle\sum_{n=1}^{\infty}\Big(\frac{h_{n}}{\lambda_{n}g_{n}}-\frac{h_{n+1}}{\lambda_{n+1}g_{n}}\Big)|A_{n}|^{2}\\
&=\displaystyle\sum_{n=1}^{\infty}\frac{h_{n}}{\lambda_{n}g_{n}}|A_{n}|^{2}-\displaystyle\sum_{n=2}^{\infty}\frac{h_{n}}{\lambda_{n}g_{n-1}}|A_{n-1}|^{2} \\
&=\displaystyle\sum_{n=1}^{\infty}\frac{h_{n}}{\lambda_{n}}\Big(\frac{|A_{n}|^{2}}{g_{n}}-\frac{|A_{n-1}|^{2}}{g_{n-1}}\Big),
\end{align*}where it is assumed that the term $\frac{|A_0|^2}{\lambda_{1}g_{0}}$ as zero. Further by assuming the term $\sqrt{\frac{g_{1}}{g_{0}}}\frac{A_{0}}{\sqrt\lambda_{1}}$ as zero, the following computation gives
\begin{align*}
&\displaystyle\sum_{n=1}^{\infty}\frac{|A_{n}-A_{n-1}|^{2}}{\lambda_{n}}-\displaystyle\sum_{n=1}^{\infty}w_{n}(\lambda,g)|A_n|^{2}\\
& =\displaystyle\sum_{n=1}^{\infty}\Big[\Big|\frac{A_{n}}{\sqrt\lambda_{n}}-\frac{A_{n-1}}{\sqrt\lambda_{n}}\Big|^{2}-\frac{h_{n}}{\lambda_{n}}\Big(\frac{|A_{n}|^{2}}{g_{n}}-\frac {|A_{n-1}|^{2}}{g_{n-1}}\Big)\Big]\\
& =\displaystyle\sum_{n=1}^{\infty}\Big[\frac{|A_{n}|^{2}}{\lambda_{n}}+\frac{|A_{n-1}|^{2}}{\lambda_{n}}-\frac{2\mathfrak{R}(\bar{A}_{n}A_{n-1})}{\lambda_{n}}-\frac{h_{n}}
{\lambda_{n}}\Big(\frac{|A_{n}|^{2}}{g_{n}}-\frac {|A_{n-1}|^{2}}{g_{n-1}}\Big)\Big]\\
&=\displaystyle\sum_{n=1}^{\infty}\Big[\Big(1-\frac{h_{n}}{g_{n}}\Big)\frac{|A_{n}|^{2}}{\lambda_{n}}+\Big(1+\frac {h_{n}}{g_{n-1}}\Big)\frac{|A_{n-1}|^{2}}{\lambda_{n}}-\frac{2\mathfrak{R}(\bar{A}_{n}A_{n-1})}{\lambda_{n}}\Big]
\end{align*}
\begin{align*}
&=\displaystyle\sum_{n=2}^{\infty}\Big|\sqrt{(1-\frac{h_{n}}{g_{n}}})\frac{A_{n}}{\sqrt{\lambda_{n}}}-\sqrt{(1+\frac{h_{n}}{g_{n-1}})}\frac{A_{n-1}}
{\sqrt\lambda_{n}}\Big|^{2}\geq0.
\end{align*}
Therefore, the above discussion gives the following identity
\begin{align}{\label{THM1IDN}}
&\displaystyle\sum_{n=2}^{\infty}\Big|\sqrt{(1-\frac{h_{n}}{g_{n}})}\frac{A_{n}}{\sqrt{\lambda_{n}}}-\sqrt{(1+\frac{h_{n}}{g_{n-1}})}
\frac{A_{n-1}}{\sqrt{\lambda_{n}}}\Big|^{2}+\displaystyle\sum_{n=1}^{\infty}\frac{h_{n}}{\lambda_{n}}\Big(\frac{|A_{n}|^{2}}{g_{n}}-\frac {|A_{n-1}|^{2}}{g_{n-1}}\Big) \nonumber\\
&=\displaystyle\sum_{n=1}^{\infty}\Big|\frac{A_{n}}{\sqrt\lambda_{n}}-\frac{A_{n-1}}{\sqrt\lambda_{n}}\Big|^{2},
\end{align}
which proves the desired inequality (\ref{GIMHI}). This completes the proof of the theorem.\\
We now prove the optimality of the sequence $w_n$. Let $\widetilde{w}_{n}$ be a sequence satisfying $\widetilde{w}_{n}\geq w_{n}$. Then using the above identity (\ref{THM1IDN}), we get
\begin{align}{\label{THM1OPTI}}
0& \leq\displaystyle\sum_{n=1}^{\infty}(\widetilde{w}_{n}-w_{n})|A_{n}|^{2}\leq\displaystyle\sum_{n=2}^{\infty}\Big|\sqrt{ \frac{g_{n-1}}{g_{n}}}\frac{A_{n}}{\sqrt\lambda_{n}}-\sqrt{\frac{g_{n}}{g_{n-1}}}\frac{A_{n-1}}{\sqrt\lambda_{n}}\Big|^{2}
\end{align}
Now choose a sequence $A_{n}=g_{n}\gamma^{N}_{n}$ in above inequality (\ref{THM1OPTI}), where $\gamma^{N}\in C_{c}(\mathbb{N}_{0})$ satisfying the given assumptions,
we get
\begin{center}
$0\leq\displaystyle\sum_{n=1}^{\infty}(\widetilde{w}_{n}-w_{n})|g_{n}\gamma^{N}_{n}|^{2}\leq\displaystyle\sum_{n=2}^{\infty}
\frac{g_{n}g_{n-1}}{\lambda_{n}}|\gamma^{N}_{n}-\gamma^{N}_{n-1}|^{2}$.
\end{center}
Since the sequence $\{g_n\}$ is strictly positive, $\gamma^{N}\leq\gamma^{N+1}$ and $\gamma^{N}\rightarrow1$ as $N\rightarrow\infty$ pointwise, so by monotone convergence theorem and (\ref{LIMCOND}), we obtain
\begin{center}
$\displaystyle\sum_{n=1}^{\infty}(\widetilde{w}_{n}-w_{n})g^{2}_{n}=0$.
\end{center}
Hence $\widetilde{w}_{n}=w_{n}$, which establishes the optimality of $w_{n}$.
\end{proof}

An immediate consequence of the Theorem \ref{THMIDHI} is the recent result of Krej\v{c}i\v{r}\'{i}k et al. (see Theorem 10, \cite{DKALFS})
\begin{cor}(Theorem 10, \cite{DKALFS})
If one chooses $\lambda_{n}=1$, $n\in \mathbb{N}$, and $\{g_n\}$ is a positive sequence of real numbers then
\begin{center}
$\displaystyle\sum_{n=1}^{\infty}w_{n}(g)|A_{n}|^2\leq\displaystyle\sum_{n=1}^{\infty}|A_{n}-A_{n-1}|^{2}$,
\end{center}
where the sequence $w_{n}=2-\frac{g_{n-1}}{g_{n}}-\frac{g_{n+1}}{g_{n}}$ is optimal if
\begin{align*}
\displaystyle\lim_{N\rightarrow\infty}\sum_{n=2}^{\infty}g_{n}g_{n-1}|\gamma^{N}_{n}-\gamma^{N}_{n-1}|^{2}=0.
\end{align*}
\end{cor}
The Theorem \ref{THMIDHI} also contains the inequality derived by Gupta \cite{SGA} as follows.
\begin{cor}(Gupta \cite{SGA})
If one chooses $\lambda_{n}=\frac{1}{n^{\alpha}}$, $\alpha\in \mathbb{R}$, and $g_n=n^{\beta}$ then
\begin{center}
$\displaystyle\sum_{n=1}^{\infty}w_{n}(\alpha,\beta)|A_{n}|^{2}$$\leq\displaystyle\sum_{n=1}^{\infty}|A_{n}-A_{n-1}|^{2}n^{\alpha}$,
\end{center}
where $\beta\in \mathbb{R}$ and
\begin{align*}
w_{n}(\alpha,\beta)&=n^{\alpha}\Big[1+(1+\frac{1}{n})^{\alpha}-(1-\frac{1}{n})^{\beta}-(1+\frac{1}{n})^{\alpha+\beta}\Big].
\end{align*}
\end{cor}
\begin{Remark}
We remark that if $w_{n}(\lambda,\beta)$ is expanded in a series then we have
\begin{align*}
&w_{n}(\lambda,\beta)\\
& =\frac{1}{\lambda_{n}}+\frac{1}{\lambda_{n+1}}-\frac{1}
{\lambda_{n}}\Big(1-\frac{1}{n}\Big)^{\beta}-\frac{1}{\lambda_{n+1}}\Big(1+\frac{1}{n}\Big)^{\beta}\\
&=\Big(\frac{1}{\lambda_{n}}-\frac{1}{\lambda_{n+1}}\Big)\frac{\beta}{n}+\frac{\beta(1-\beta)}{2}\frac{1}{n^{2}}\Big(\frac{1}{\lambda_{n}}+\frac{1}{\lambda_{n+1}}\Big)+ \frac{\beta(\beta-1)(\beta-2)}{6}\frac{1}{n^{3}}\Big(\frac{1}{\lambda_{n}}-\frac{1}{\lambda_{n+1}}\Big)\\
& \hspace{0.15cm}
+\frac{\beta(1-\beta)(\beta-2)(\beta-3)}{24}\frac{1}{n^{4}}\Big(\frac{1}{\lambda_{n}}+\frac{1}{\lambda_{n+1}}\Big)+\cdots\cdots
\end{align*}
Hence if $\beta\in(0,1)$ and the sequence $\{\lambda_{n}\}$ is non-decreasing then one gets
\begin{center}
$w_{n}(\lambda,\beta)>\frac{\beta(1-\beta)}{2}\frac{1}{n^{2}}(\frac{1}{\lambda_{n}}+\frac{1}{\lambda_{n+1}})$.
\end{center}
\end{Remark}
Therefore for any $\beta\in(0,1)$ and $\lambda_{n+1}\geq\lambda_{n}$, $n\in\mathbb{N}$ one can asserts the following:
\begin{align}{\label{RBETA}}
\displaystyle\sum_{n=1}^{\infty}\Big(\frac{1}{\lambda_{n}}+\frac{1}{\lambda_{n+1}}\Big)\frac{\beta(1-\beta)}{2}\frac{1}{n^{2}}|A_{n}|^{2}
&<\displaystyle\sum_{n=1}^{\infty}w_{n}(\lambda,\beta) |A_{n}|^{2}\leq\displaystyle\sum_{n=1}^{\infty}\frac{|A_{n}-A_{n-1}|^{2}}{\lambda_{n}}.
\end{align}

\begin{Remark}{\label{RGIMHI}}
In particular, if $\beta=\frac{1}{2}$ is chosen in $w_{n}(\lambda,\beta)$, then we get
$$w_{n}(\lambda, \frac{1}{2})=\frac{1}{\lambda_{n}}+\frac{1}{\lambda_{n+1}}-\frac{1}
{\lambda_{n}}\sqrt{(1-\frac{1}{n})}-\frac{1}{\lambda_{n+1}}\sqrt{(1+\frac{1}{n})}.$$
Moreover, we note that
\begin{center}
$w_{n}(\lambda, \frac{1}{2})>\frac{1}{8n^{2}}(\frac{1}{\lambda_{n}}+\frac{1}{\lambda_{n+1}})$
\end{center}
is true for all non-decreasing sequence $\{\lambda_{n}\}$.
\end{Remark}

\begin{cor}
It is to be noted that for $\lambda_{n}=1$, $n\in \mathbb{N}$ and $\beta=\frac{1}{2}$ then we get the following inequality due to Keller et al. \cite{MKR}:
\begin{center}
$\displaystyle\sum_{n=1}^{\infty}\frac{1}{4n^{2}}|A_{n}|^{2} < \displaystyle\sum_{n=1}^{\infty}w_{n}|A_{n}|^{2} \leq \displaystyle\sum_{n=1}^{\infty}|A_{n}-A_{n-1}|^{2}$
\end{center}
where $ w_{n}\equiv w_{n}(\{1\}, \frac{1}{2})=2-\sqrt{(1-\frac{1}{n})}-\sqrt{(1+\frac{1}{n})}$.
\end{cor}
Therefore, from the above discussions, we have the following theorem.
\begin{thm}{\label{THMOPTI}} Let $g_{n}=n^{\beta}$, $\beta\in (0, 1)$ and $\lambda=\{\lambda_{n}\}$ be a non-decreasing sequence of real numbers such that $\lambda_n>0$. Then for any sequence $A=\{A_n\}$  of real or complex numbers with $A\in C_c(\mathbb{N}_0)$ such that $A_{0}=0$ the following inequality holds:
\begin{align}
\displaystyle\sum_{n=1}^{\infty}\Big(\frac{1}{\lambda_{n}}+\frac{1}{\lambda_{n+1}}\Big)\frac{\beta(1-\beta)}{2n^{2}}|A_{n}|^{2}
& < \displaystyle\sum_{n=1}^{\infty}w_{n}(\lambda,\beta)|A_{n}|^{2} \leq \displaystyle\sum_{n=1}^{\infty}\frac{|A_{n}-A_{n-1}|^{2}}{\lambda_{n}}.
\end{align}Also for $\lambda_n\geq 1$, $n\in \mathbb{N}$, and $\beta\in (0,\frac{1}{2}]$ the weight sequence $w_{n}(\lambda, \beta)$ is optimal.
\end{thm}
\begin{proof}
The proof of the theorem is a combination of Theorem \ref{THMIDHI} and inequality (\ref{RBETA}). We now only establish the optimality of the weight sequence $w_{n}(\lambda,\beta)$ here, and for this we follow the  Krej\v{c}i\v{r}\'{i}k and \v{S}tampach \cite{DKK} approach. Note that for $\beta\in (0,\frac{1}{2}]$ the identity (\ref{THM1IDN}) gives
\begin{align}{\label{GHI}}
\displaystyle\sum_{n=1}^{\infty}\frac{|A_{n}-A_{n-1}|^{2}}{\lambda_{n}}
&=\displaystyle\sum_{n=2}^{\infty}\Big|\Big(\frac{n-1}{n}\Big)^{\frac{\beta}{2}}\frac{A_n}{\sqrt{\lambda_{n}}}-\Big(\frac{n}{n-1}\Big)^{\frac{\beta}{2}}\frac{A_{n-1}}{\sqrt{\lambda_{n}}}\Big|^{2}
+\displaystyle\sum_{n=1}^{\infty}w_{n}(\lambda,\beta)|A_{n}|^{2}.
\end{align}
Suppose $\widetilde{w_{n}}(\lambda,\beta)$ is a sequence such that $\widetilde{w_{n}}(\lambda,\beta)\geq w_{n}(\lambda,\beta)$ holds for all $n\in \mathbb{N}$ and inequality (\ref{GHI}) holds for the sequence $\widetilde{w_{n}}(\lambda,\beta)$. Then for all $A\in C_c(\mathbb{N}_0)$, we have the following:
\begin{align}{\label{WNTILDE}}
\displaystyle\sum_{n=2}^{\infty}
\Big|\Big(\frac{n-1}{n}\Big)^{\frac{\beta}{2}}\frac{A_n}{\sqrt{\lambda_{n}}}-\Big(\frac{n}{n-1}\Big)^{\frac{\beta}{2}}\frac{A_{n-1}}{\sqrt{\lambda_{n}}}\Big|^{2}& \geq \displaystyle\sum_{n=1}^{\infty}(\widetilde{w_{n}}(\lambda,\beta)-w_{n}(\lambda,\beta))|A_{n}|^{2} \geq 0.
\end{align}
Since inequality (\ref{WNTILDE}) vanishes when $A_n=n^{\beta}$, $ \beta\in (0,\frac{1}{2}]$ which is not a finitely supported sequence, so we redefine the sequence $A^{N}_{n}$, $N\geq2$ as $A^{N}_{n}=\gamma^{N}_{n}n^{\beta}$ where $\gamma^{N}_{n}$ defined as below:
\begin{align*}
\gamma^{N}_{n} &= \left\{
\begin{array}{lll}
    1 & \quad \mbox{if~~} n<N,\\
    \frac {2\log N-\sqrt\lambda_{n}\log n}{\log N} & \quad \mbox{if~~} N\leq n\leq N^{2}\\
    0 & \quad \mbox{if~~} n>N^{2}.
\end{array}\right.
\end{align*}
Now
\begin{align*}
&\displaystyle\sum_{n=2}^{\infty}\Big|\Big(\frac{n-1}{n}\Big)^{\frac{\beta}{2}}\frac{A_n^N}{\sqrt{\lambda_{n}}}-\Big(\frac{n}{n-1}\Big)^{\frac{\beta}{2}}\frac{A_{n-1}^N}{\sqrt{\lambda_{n}}}\Big|^{2}\\
&=\displaystyle\sum_{n=2}^{\infty}\frac{1}{\lambda_n}\big(n(n-1)\big)^{\beta}\Big|\gamma^{N}_{n}-\gamma^{N}_{n-1}\Big|^{2}\\
&=\frac{1}{(\log N)^{2}}\displaystyle\sum_{n=N+1}^{N^{2}}\frac{1}{\lambda_{n}}\big(n(n-1)\big)^{\beta}\Big|\sqrt{\lambda_{n-1}}\log(n-1)-\sqrt{\lambda_{n}}\log n\Big|^{2}\\
&\leq \frac{1}{(\log N)^{2}}\displaystyle\sum_{n=N+1}^{N^{2}}\sqrt{n(n-1)}\Big(\log\Big(\frac{n}{n-1}\Big)\Big)^2~~~ (\mbox{since~} \{\lambda_{n}\} \mbox{~is non-decreasing})\\
& \leq\frac{1}{(\log N)^{2}}\displaystyle\sum_{n=N+1}^{N^{2}}\frac{\sqrt{n(n-1)}}{(n-1)^{2}} \leq \frac{1}{(\log N)^{2}}\displaystyle\sum_{n=N+1}^{N^{2}}\Big\{\frac{1}{(n-1)}+\frac{1}{2(n-1)^2}\Big\}
\end{align*}
\begin{align*}
& \leq \frac{1}{(\log N)^{2}} \displaystyle\int_{N}^{N^2}\Big\{\frac{1}{(n-1)}+\frac{1}{2(n-1)^2}\Big\}dn\\
& \leq \frac{1}{\log N} + \frac{N}{2(N^2-1)(\log N)^{2}},
\end{align*}
which tends to $0$ as $N\rightarrow\infty$. Hence from (\ref{WNTILDE}), we get
\begin{center}
$\displaystyle\sum_{n=1}^{\infty}(\widetilde{w_{n}}(\lambda,\beta)-w_{n}(\lambda,\beta))|A_{n}|^{2}=0$.
\end{center}
Therefore, $\widetilde{w_{n}}(\lambda,\beta)= w_{n}(\lambda,\beta)$ holds for all $n\in \mathbb{N}$, that is the sequence $w_{n}(\lambda,\beta)$, $\beta\in(0,\frac{1}{2}]$ is optimal. This completes proof of the theorem.
\end{proof}
\section{Improvement of Copson's inequality}
In this section, we present an improvement of Copson's inequality (\ref{COPIQNEQN}). To achieve it, first we need to establish the following theorem, the statement of which is given below.
\begin{thm}
Let $\{a_n\}$ be any sequence of real or complex numbers such that $A_{n}=\displaystyle\sum_{k=1}^{n}ka_{k}$ with $A_{0}=0$. Then for $1<c\leq 2$, we have
\begin{align}{\label{ineqimvd}}
\displaystyle\sum_{n=1}^{\infty}V_{n}|A_n|^{2}&\leq\displaystyle\sum_{n=1}^{\infty}
S_n^{2-c}\frac{|A_{n}-A_{n-1}|^{2}}{n},
\end{align}where the sequence $V_{n}$ is defined as below:
\begin{align*}
V_{n}& =\frac{S_n^{2-c}}{n}+\frac{S_{n+1}^{2-c}}{n+1}-\frac{S_n^{2-c}}{n}\sqrt{1-\frac{1}{n}}-\frac{S_{n+1}^{2-c}}{n+1}\sqrt{1+\frac{1}{n}},
\end{align*} with $S_n=\frac{n(n+1)}{2}$.
\end{thm}
\begin{proof}
Let us denote $\tau_{n}=\frac{\sqrt n-\sqrt{ (n-1)}}{n}$ so that $\tau_{n+1}=\frac{\sqrt{(n+1)}-\sqrt n}{n+1}$. Then $V_{n}$ reduces to the following form:
\begin{center}
$V_{n}= \frac{1}{\sqrt n}\Big(\tau_{n}S_n^{2-c}-\tau_{n+1}S_{n+1}^{2-c}\Big)$.
\end{center}
Thus using the above expression for $V_n$, one can have the following identity.
\begin{align*}
\displaystyle\sum_{n=1}^{\infty}V_{n}|A_{n}|^{2}
& =\displaystyle\sum_{n=1}^{\infty}\Big[\frac{\tau_{n}S_n^{2-c}}{\sqrt n}-\frac{\tau_{n+1}S_{n+1}^{2-c}}{\sqrt n}\Big]|A_{n}|^{2}\\
&=\displaystyle\sum_{n=1}^{\infty}\frac{\tau_{n}}{\sqrt n}S_n^{2-c}|A_{n}|^{2}-\displaystyle\sum_{n=2}^{\infty}
\frac{\tau_{n}}{\sqrt{(n-1)}}S_n^{2-c}|A_{n-1|^{2}}\\
&=\displaystyle\sum_{n=1}^{\infty}\tau_{n}\Big(\frac{|A_{n}|^{2}}{\sqrt n}-\frac{|A_{n-1}|^{2}}{\sqrt{(n-1)}}\Big)S_n^{2-c}.
\end{align*}
Now let us compute the difference between the terms in both sides of inequality (\ref{ineqimvd}), that is
\begin{align}{\label{COPMAINREI}}
&\displaystyle\sum_{n=1}^{\infty}
\frac{|A_{n}-A_{n-1}|^{2}}{n}S_n^{2-c}-\displaystyle\sum_{n=1}^{\infty}V_{n}|A_n|^{2} \nonumber\\
& =\displaystyle\sum_{n=1}^{\infty}\Big[\Big|\frac{A_{n}}{\sqrt n}-\frac{A_{n-1}}{\sqrt n}\Big|^{2}S_n^{2-c}-
\tau_{n}\Big(\frac{|A_{n}|^{2}}{\sqrt n}-\frac {|A_{n-1}|^{2}}{\sqrt{(n-1)}}\Big)S_n^{2-c}\Big] \nonumber\\
& =\displaystyle\sum_{n=1}^{\infty}\Big[\frac{|A_{n}|^{2}}{n}+\frac{|A_{n-1}|^{2}}{n}-\frac{2\mathfrak{R}(\bar{A}_{n}A_{n-1})}{n}-
\tau_{n}\Big(\frac{|A_{n}|^{2}}{\sqrt n}-\frac {|A_{n-1}|^{2}}{\sqrt{(n-1)}}\Big)\Big]S_n^{2-c} \nonumber\\
&=\displaystyle\sum_{n=1}^{\infty}\Big[\Big(1-\frac{n\tau_{n}}{\sqrt n}\Big)
\frac{|A_{n}|^{2}}{n}+\Big(1+\frac {n\tau_{n}}{\sqrt{(n-1)}}\Big)\frac{|A_{n-1}|^{2}}{n}-\frac{2\mathfrak{R}(\bar{A}_{n}A_{n-1})}{n}\Big]S_n^{2-c} \nonumber\\
&=\displaystyle\sum_{n=1}^{\infty}\Big|\sqrt{(1-\frac{n\tau_{n}}{\sqrt n})}\frac{A_{n}}{\sqrt n}
-\sqrt{(1+\frac{n\tau_{n}}{\sqrt{(n-1)}})}\frac{A_{n-1}}{\sqrt{n}}
\Big|^{2}S_n^{2-c} \nonumber \\
&=\displaystyle\sum_{n=1}^{\infty}\Big|\frac{A_{n}}{\sqrt n}\sqrt[4]{\frac{n-1}{n}}
-\frac{A_{n-1}}{\sqrt n}\sqrt[4]{\frac{n}{n-1}}\Big|^{2} S_n^{2-c}\\
& \geq 0. \nonumber
\end{align}
Hence the desired inequality (\ref{ineqimvd}) follows easily.
\end{proof}

Next we present few results with the aim of establishing improvement of Copson's inequality (\ref{COPIQNEQN}).
\begin{Lemma}{\label{COPIML1}}
If $2\geq c\geq\frac{3}{2}$ then $\forall n\in\mathbb{N}$ we get
\begin{align*}
\frac{S_n^{2-c}}{n}& >\frac{S_{n+1}^{2-c}}{n+1}.
\end{align*}
\end{Lemma}
\begin{proof}We know that for $n\in \mathbb{N}$, $(n+1)^{2}>n^{2}+2n$. Hence we have
$\Big(\frac{n+1}{n}\Big)^{2}>\frac{n+2}{n}$.\\
Therefore for each $c\geq\frac{3}{2}$, we get
\begin{align*}
\frac{n+1}{n}& >\Big(\frac{n+2}{n}\Big)^{\frac{1}{2}}\geq\Big(\frac{n+2}{n}\Big)^{2-c}.
\end{align*}
Now an easy computation gives
\begin{align*}
\frac{S_n^{2-c}}{n}-\frac{S_{n+1}^{2-c}}{n+1}
& > \Big(\frac{n+2}{n}\Big)^{2-c}\frac{S_n^{2-c}}{n+1}-\frac{S_{n+1}^{2-c}}{n+1}=0,
\end{align*}which proves the desired inequality.
\end{proof}

\begin{Lemma}{\label{COPIML2}}
For any $2\geq c \geq \frac{3}{2}$, we get
\begin{align*}
V_{n}&>\Big(\frac{S_n^{2-c}}{n}+\frac{S_{n+1}^{2-c}}{n+1}\Big)\frac{1}{8n^{2}}.
\end{align*}
\end{Lemma}
\begin{proof}
Using the Lemma \ref{COPIML1}, one can deduce the following inequality:
\begin{align*}
V_{n}&=\frac{S_n^{2-c}}{n}+\frac{S_{n+1}^{2-c}}{n+1}-\frac{S_n^{2-c}}{n}\sqrt{(1-\frac{1}{n})}-\frac{S_{n+1}^{2-c}}{n+1}\sqrt{(1+\frac{1}{n})}\\
&=\Big(\frac{S_n^{2-c}}{n}-\frac{S_{n+1}^{2-c}}{n+1}\Big)\frac{1}{2n}+\Big(\frac{S_n^{2-c}}{n}+\frac{S_{n+1}^{2-c}}{n+1}\Big)\frac{1}{8n^{2}}+
\Big(\frac{S_n^{2-c}}{n}-\frac{S_{n+1}^{2-c}}{n+1}\Big)\frac{1}{16n^{3}}+\Big(\frac{S_n^{2-c}}{n}\\
& \hspace{1cm}+\frac{S_{n+1}^{2-c}}{n+1}\Big)\frac{5}{128n^{4}}+\cdots\cdots\\
&>\Big(\frac{S_n^{2-c}}{n}+\frac{S_{n+1}^{2-c}}{n+1}\Big)\frac{1}{8n^{2}}+\Big(\frac{S_n^{2-c}}{n}+\frac{S_{n+1}^{2-c}}{n+1}\Big)\frac{5}{128n^{4}}+\cdots\cdots
\end{align*}Hence we have desired inequality.
\end{proof}

\begin{Lemma}{\label{COPIML3}}
Let $n\in \mathbb{N}$. Then for any $1<c\leq\frac{3}{2}$, we have
\begin{align*}
\frac{1}{8n^{2}}\Big[\frac{S_n^{2-c}}{n}+\frac{S_{n+1}^{2-c}}{n+1}\Big]& > \frac{(c-1)^{2}}{4}\frac{n}{S_n^{c}}.
\end{align*}
\end{Lemma}
\begin{proof}
Let $n\in \mathbb{N}$ and $c>1$. Then we have the following steps
\begin{align*}
& c\leq\frac{3}{2}\\
\Rightarrow &8(c-1)^{2}\leq2<(1+\frac{1}{n})^{2}+(1+\frac{1}{n})(1+\frac{2}{n})^{2-c}\\
\Rightarrow & 8(c-1)^{2}<\frac{(n+1)^{2}}{n^{2}}+(n+1)(n+2)^{2-c}n^{c-3}\\
\Rightarrow &\frac {n(c-1)^{2}}{4}<\Big[\frac{(n+1)^{2}n}{4}+\frac{(n+1)(n+2)^{2-c}}{4}n^{c}\Big]\frac{1}{8n^{2}}=\Big[\frac{S_n^{2}}{n}+\frac{S_n^{c}S_{n+1}^{2-c}}{n+1}\Big]\frac{1}{8n^{2}}\\
\Rightarrow&\frac{(c-1)^{2}}{4}\frac{n}{S_n^{c}}<\frac{1}{8n^{2}}\Big[\frac{S_n^{2-c}}{n}+\frac{S_{n+1}^{2-c}}{n+1}\Big].
\end{align*}
Hence the desired result.
\end{proof}

\begin{Lemma}{\label{COPIML4}}
Suppose that $n\in \mathbb{N}$. Then we have
\begin{align*}
V_n& >\frac{1}{8n^{2}}\Big[\frac{\sqrt{S_n}}{n}+\frac{\sqrt{S_{n+1}}}{n+1}\Big]> \frac{1}{16}\frac{n}{{S_n\sqrt{S_n}}}.
\end{align*}
\end{Lemma}
\begin{proof}
Since Lemma \ref{COPIML2} holds for all $\frac{3}{2}\leq c \leq 2$, and Lemma \ref{COPIML3} satisfied for $1<c\leq\frac{3}{2}$, so at the intersecting point that is for $c=\frac{3}{2}$ both the Lemmas hold. Therefore, by combining both these Lemma for the case when $c=\frac{3}{2}$, we get the desired inequality.
\end{proof}
Therefore using the Lemma \ref{COPIML4}, we get an improvement of the inequality (\ref{COPIQNEQN}) for the case when $c=\frac{3}{2}$ as follows:
\begin{align}{\label{IMCOP}}
\displaystyle \sum_{n=1}^{\infty}\frac{1}{16}\frac{n}{{S_n\sqrt{S_n}}}|A_{n}|^{2} & <\displaystyle\sum_{n=1}^{\infty}V_{n}|A_{n}|^{2}\leq\displaystyle \sum_{n=1}^{\infty}{\sqrt{S_n}}\frac{|A_{n}-A_{n-1}|^{2}}{n}.
\end{align}
Now we prove the optimality of the sequence $V_n$. For this let $\widetilde{V_{n}}$ is a sequence such that $\widetilde{V_{n}}\geq V_{n}$ holds for all $n\in \mathbb{N}$ and inequality (\ref{COPMAINREI}) holds for the sequence $\widetilde{V_{n}}$. Then we have the following:
\begin{align}{\label{COPREM}}
\displaystyle\sum_{n=2}^{\infty}
\Big|\sqrt[4]{\frac{n-1}{n}}\frac{A_n}{\sqrt n}-\sqrt[4]{\frac{n}{n-1}}\frac{A_{n-1}}{{\sqrt n}}\Big|^{2}\sqrt{S_n} &\geq \displaystyle\sum_{n=1}^{\infty}(\widetilde{V_{n}}-V_{n})|A_{n}|^{2} \geq 0.
\end{align}
We now redefine the sequence $A^{N}_{n}$, $N\geq2$ as $A^{N}_{n}=\gamma^{N}_{n}\sqrt{n}$ where $\gamma^{N}_{n}$ defined as below:
\begin{align*}
\gamma^{N}_{n} &= \left\{
\begin{array}{lll}
    1 & \quad \mbox{if~~} n<N,\\
    \frac {2\log N- \sqrt[4]{\frac{2n}{(n+1)}}\log n}{\log N} & \quad \mbox{if~~} N\leq n\leq N^{2}\\
    0 & \quad \mbox{if~~} n>N^{2}.
\end{array}\right.
\end{align*}
Now using the above sequence, and Theorem \ref{THMOPTI} at the last stage, we obtain
\begin{align*}
&\displaystyle\sum_{n=2}^{\infty}\Big|\sqrt[4]{\frac{n-1}{n}}A_{n}^N-\sqrt[4]{\frac{n}{n-1}}A_{n-1}^N\Big|^{2}{\sqrt\frac{(n+1)}{2n}}\\
&=\displaystyle\sum_{n=2}^{\infty}\Big|\sqrt[4]{\frac{n-1}{n}}\gamma^{N}_{n}\sqrt{n}-\sqrt[4]{\frac{n}{n-1}}\gamma^{N}_{n-1}\sqrt{(n-1)}\Big|^{2}{\sqrt\frac{(n+1)}{2n}}\\
&=\displaystyle\sum_{n=2}^{\infty}\sqrt{n(n-1)}\Big|\gamma^{N}_{n}-\gamma^{N}_{n-1}\Big|^{2}{\sqrt\frac{(n+1)}{2n}}\\
&=\frac{1}{(\log N)^{2}}\displaystyle\sum_{n=N+1}^{N^{2}}\sqrt{n(n-1)}\Big|\sqrt[4]{\frac{2(n-1)}{n}}\log(n-1)-\sqrt[4]{\frac{2n}{n+1}}\log n\Big|^{2}{\sqrt\frac{(n+1)}{2n}}\\
&\leq\frac{1}{(\log N)^{2}}\displaystyle\sum_{n=N+1}^{N^{2}}\sqrt{n(n-1)}\Big|\sqrt[4]{\frac{2(n-1)}{n}}\log(n-1)-\sqrt[4]{\frac{2(n-1)}{n}}\log(n)\Big|^{2}{\sqrt\frac{(n+1)}{2n}}\\
& \hspace{0.5cm}\big(\mbox{since~} \sqrt[4]{\frac{2(n-1)}{n}}\leq \sqrt[4]{\frac{2n}{n+1}} ~~\&~~ \sqrt {1-\frac{1}{n^{2}}}\leq1\big)\\
&\leq \frac{1}{(\log N)^{2}}\displaystyle\sum_{n=N+1}^{N^{2}}\sqrt{n(n-1)}(\log(\frac{n}{n-1}))^2\\
& \leq\frac{1}{(\log N)^{2}}\displaystyle\sum_{n=N+1}^{N^{2}}\frac{\sqrt{n(n-1)}}{(n-1)^{2}}\\
& \leq \frac{1}{\log N} + \frac{N}{2(N^2-1)(\log N)^{2}},
\end{align*}
which tends to $0$ as $N\rightarrow\infty$. Hence from (\ref{COPREM}), we get
\begin{center}
$\displaystyle\sum_{n=1}^{\infty}(\widetilde{V}_n-V_n)|A_{n}|^{2}=0$.
\end{center}
Therefore, $\widetilde{V}_n= V_n$ holds for all $n\in \mathbb{N}$, hence the sequence $V_{n}$ is optimal. This finishes proof of the optimality.

\section{Fundamental structure of sequence space $\Gamma_p$}
In an unpublished preprint of Fischer et al. \cite{FFR}, an improved Hardy's inequality is obtained for a general $p>1$. Indeed, for $p>1$ it was proved that
\begin{align}{\label{IHGP}}
\displaystyle\sum_{n=1}^{\infty}|A_n-A_{n-1}|^p \geq \sum_{n=1}^{\infty}w_n|A_n|^p>\Big(\frac{p-1}{p}\Big)^p\sum_{n=1}^{\infty}\frac{1}{n^p}|A_n|^p,
\end{align}where $w_n=\Big(1-\Big(\frac{n-1}{n}\Big)^{\frac{p-1}{p}}\Big)^{p-1}-\Big(\Big(\frac{n+1}{n}\Big)^{\frac{p-1}{p}}-1\Big)^{p-1}$, and $A_n=a_1+a_2+\ldots+a_n$. Note that in the case of $p=2$, one gets the inequality (\ref{IMHI2}), an improved version of discrete Hardy's inequality by Keller et al. \cite{MKR}. Therefore, it is natural to study some fundamental properties of the underlying space $\Gamma_p$, which is defined as follows. Let $s$ be denoted as the space of all real or complex sequences, $\{\gamma_n\}$, and $\{q_n\}$ be two sequences of real numbers such that $\gamma_n>0$, and $q_n>0$ for each $n\in \mathbb{N}$. Then for $1<p<\infty$, we define
\begin{align}{\label{GPSEQ}}
\Gamma_p&= \Big\{x=\{a_n\}\in s: \displaystyle\sum_{n=1}^{\infty} \gamma_n\Big|\sum_{k=1}^{n}q_ka_k\Big|^p < \infty\Big\},
\end{align} and when $\gamma_n=\frac{1}{n^p}$ with $q_n=1$ for all $n\in \mathbb{N}$, we denote $\Gamma_p$ by $X_p$, where
\begin{align}{\label{XPSEQ}}
X_p&= \Big\{x=\{a_n\}\in s: \displaystyle\sum_{n=1}^{\infty} \frac{1}{n^p}\Big|\sum_{k=1}^{n}a_k\Big|^p < \infty\Big\}.
\end{align}
The sequence space $X_p$ is known as Ces\`{a}ro sequence spaces of non-absolute type and studied in \cite{PNNGLEE}. Note that when we choose $\gamma_n=w_n$, and $q_n=1$ for all $n\in \mathbb{N}$ in (\ref{GPSEQ}), then $\Gamma_p$ reduces to the sequence space $W_p$ (say), which is directly involved in the inequality (\ref{IHGP}). In case when $p=2$, sequence space $W_2$ is connected with (\ref{IMHI2}). Also for $\gamma_n=V_n$, $p=2$, and $q_n=n$ for all $n\in \mathbb{N}$, we get a sequence space, which is demonstrated in (\ref{IMCOP}).
It is easy to verify that the sequence space $\Gamma_p$ defined as in (\ref{GPSEQ}) is a normed linear space equipped with the norm functional $\|\cdot\|_{\Gamma_p}$ defined as below:
\begin{center}
$\|x\|_{\Gamma_p}=\Big(\displaystyle\sum_{n=1}^{\infty}\gamma_{n} \Big|\displaystyle\sum_{k=1}^{n}q_ka_{k}\Big|^{p}\Big)^{\frac{1}{p}}$.
\end{center}
The sequence space $\Gamma_p$ is of non-absolute type, which means the absolute value of the sequence $|x|=\{|a_n|\}_{n\geq1}$ doesn't belong to $\Gamma_p$ whenever $x\in \Gamma_p$. The following result demonstrates the completeness of the space $\Gamma_p$. Therefore, let us begin with the following proposition.
\begin{Proposition}
The pair $(\Gamma_p, \|\cdot\|_{\Gamma_p})$ is a Banach sequence space of non-absolute type.
\end{Proposition}
\begin{proof}
It is only required to show the completeness of the space $(\Gamma_p, \|\cdot\|_{\Gamma_p})$. Suppose that $\{x^{(i)}\}_{i\geq 1}$, where $x^{(i)}=\{a_n^{(i)}\}_{n\geq 1}$ is a cauchy sequence in $(\Gamma_p, \|\cdot\|_{\Gamma_p})$. Then for any $\varepsilon>0$ there exists a natural number $N_0$ such that $\|x^{(i)}-x^{(j)}\|_{\Gamma_p}<\varepsilon$ for all $i, j>N_0$. Hence by definition, we have
\begin{center}
$\|x^{(i)}-x^{(j)}\|_{\Gamma_p}^{p}=\Big(\displaystyle\sum_{n=1}^{\infty}\gamma_{n} \Big|\displaystyle\sum_{k=1}^{n}q_k(a_{k}^{(i)}-a_{k}^{(j)})\Big|^{p}\Big)<\varepsilon^p$.
\end{center}
Since $\gamma_n>0$, and $q_n>0$ for each $n\in \mathbb{N}$, so one can easily concludes that for each $n\in \mathbb{N}$ the sequence $\{a_n^{(i)}\}_{i\geq 1}$ is Cauchy in $\mathbb{C}$, hence converges. Let $\displaystyle\lim_{j\rightarrow\infty}a_n^{(j)}=a_n$ for a fixed $n\in \mathbb{N}$, where $x=\{a_n\}_{n\geq 1}$. Now fix any $m\in \mathbb{N}$. For a given $\varepsilon>0$, there exists $N_0\in \mathbb{N}$, we have
\begin{center}
$\Big(\displaystyle\sum_{n=1}^{m}\gamma_{n} \Big|\displaystyle\sum_{k=1}^{n}q_k(a_{k}^{(i)}-a_{k})\Big|^{p}\Big)^{\frac{1}{p}}<\varepsilon$.
\end{center}Choosing $m\rightarrow\infty$, we get $\|x^{(i)}-x\|_{\Gamma_p}<\varepsilon$ for all $i>N_0$. This completes the proof. Therefore, $(\Gamma_p, \|\cdot\|_{\Gamma_p})$ is a Banach sequence space of non-absolute type.
\end{proof}
Let us define an infinite matrix $G=(g_{nk})_{n, k\geq 1}$ as below:
 \begin{align*}
g_{nk} &= \left\{
\begin{array}{ll}
    \gamma_n^{\frac{1}{p}}q_k & \quad 1\leq k \leq n,\\ \vspace{0.2cm}
    0 & \quad k>n.
\end{array}\right.
\end{align*}
Note that the matrix $G$ includes more general matrices such as generalized Ces\`{a}ro matrix $C^N$ when one chooses $q_k=1$, and $\gamma_n=\frac{1}{(n+N)^p}$, weighted mean matrix when $\gamma_n=\frac{1}{Q_n^p}$, and etc. These matrices are studied by Chen et al. \cite{CHENLOUR} in connection with Hardy's inequality. Further, it is easy to observe that the sequence space $\Gamma_p$ can be written equivalently as $\Gamma_p=\big\{x=\{a_n\}\in s: ~Gx \in l_p\big\}$. Also since $G$ is lower triangular, so it is invertible, and the inverse matrix $G^{-1}$ is defined as below:
\begin{align*}
g_{nk}^{-1} &= \left\{
\begin{array}{ll}
    \frac{\gamma_k^{-\frac{1}{p}}}{q_n} & \quad k = n, n\geq 1,\\ \vspace{0.2cm}
    -\frac{\gamma_{k}^{-\frac{1}{p}}}{q_n} & \quad k = n-1, n\geq 2,\\
    0 & \quad \mbox{elsewhere}.
\end{array}\right.
\end{align*}
Now we have another proposition given below.
\begin{Proposition}
The space $(\Gamma_p, \|\cdot\|_{\Gamma_p})$ is linearly isomorphic to the $p$-summable sequence space $(l_p, \|\cdot\|_{p})$.
\end{Proposition}
\begin{proof}
We need to prove that there is an existence of a bijective and norm preserving linear map $T:\Gamma_p\rightarrow l_p$. Using the matrix $G$, we define $T$ as $Tx=Gx$ for all $x\in \Gamma_p$. Then $T$ is linear, and for any $y=\{b_n\}_{n\geq 1}\in l_p$, with $b_0=0$ we have $y=Gx$, and hence $x=G^{-1}y$. In fact the terms of the sequence $x=\{a_n\}_{n\geq 1}$ are $a_n=\frac{1}{q_n}(\gamma_n^{-\frac{1}{p}}b_n-\gamma_{n-1}^{-\frac{1}{p}}b_{n-1})$, for $n\in \mathbb{N}$. The map $T$ is also norm preserving as
\begin{align*}
\|x\|_{\Gamma_p}^p&=\displaystyle\sum_{n=1}^{\infty}\gamma_{n} \Big|\displaystyle\sum_{k=1}^{n}q_ka_{k}\Big|^{p}=\displaystyle\sum_{n=1}^{\infty}\gamma_{n} \Big|\displaystyle\sum_{k=1}^{n}\gamma_k^{-\frac{1}{p}}b_k-\gamma_{k-1}^{-\frac{1}{p}}b_{k-1}\Big|^{p}= \sum_{n=1}^{\infty}|b_n|^p=\|y\|_p^p=\|Tx\|_p^p.
\end{align*}One can also easily established that the map $T$ is onto. Therefore, we conclude that there exists a isometric isomorphism between sequence spaces $\Gamma_p$ and $l_p$. This completes the proof.
\end{proof}

\begin{Proposition}
The space $(\Gamma_p, \|\cdot\|_{\Gamma_p})$ is not a Hilbert space except when $p\neq 2$.
\end{Proposition}
\begin{proof}
In case when $p=2$, then it is very easy to verify that the parallelogram identity $\|x+y\|_{\Gamma_p}^2+\|x-y\|_{\Gamma_p}^2=2(\|x\|_{\Gamma_p}^2+\|y\|_{\Gamma_p}^2)$ holds. Hence $\Gamma_2$ is a Hilbert space. Now choose $p\neq 2$, and $x=\{\frac{\gamma_1^{-\frac{1}{p}}}{q_1}, \frac{1}{q_2}(\gamma_2^{-\frac{1}{p}}-\gamma_1^{-\frac{1}{p}}), -\frac{\gamma_2^{-\frac{1}{p}}}{q_3}, 0, 0, \ldots\}$, $y=\{\frac{\gamma_1^{-\frac{1}{p}}}{q_1}, -\frac{1}{q_2}(\gamma_2^{-\frac{1}{p}}+\gamma_1^{-\frac{1}{p}}), \frac{\gamma_2^{-\frac{1}{p}}}{q_3}, 0, 0, \ldots\}$. For such sequences, we have $\|x+y\|_{\Gamma_p}^2=4$, $\|x-y\|_{\Gamma_p}^2=4$, $\|x\|_{\Gamma_p}^2=4^{1/p}=\|y\|_{\Gamma_p}^2$. Therefore, $\|x+y\|_{\Gamma_p}^2+\|x-y\|_{\Gamma_p}^2=8\neq 4^{\frac{p+1}{p}}=2(\|x\|_{\Gamma_p}^2+\|y\|_{\Gamma_p}^2)$ is true for $p\neq 2$. Hence the result follows.
\end{proof}

\begin{Proposition}
The following inclusions results are true:\\
$(a)$ if $\gamma_n=w_n$, $q_n=1$ for all $n\in \mathbb{N}$ then for any $p>1$, $l_p\subset W_p \subseteq X_p$.\\
$(b)$ if $\{Q_n\gamma_n^{\frac{1}{p}}\}_{n\geq 1}\in l_p$ then $l_\infty\subset \Gamma_p$.
\end{Proposition}
\begin{proof}
$(a)$ The result directly follows from the inequality (\ref{IHGP}). Note that the series $\displaystyle\sum_{n=1}^{\infty}w_n$ converges. The inclusion $l_p\subset W_p$ is strict as the sequence $x=\{(-1)^n\}_{n\geq 1}\in W_p$ but $x\notin l_p$.\\
$(b)$ Since $x\in l_\infty$, so for each $n\in \mathbb{N}$, there is a constant $M>0$ such that $\Big|\displaystyle\sum_{k=1}^{n}q_ka_{k}\Big|^p\leq (MQ_n)^p$, where $Q_n=q_1+\ldots+q_n>0$. Hence the desired inclusion follows easily from $\displaystyle\sum_{n=1}^{\infty} \gamma_n\Big|\sum_{k=1}^{n}q_ka_k\Big|^p \leq \sum_{n=1}^{\infty}Q_n^p\gamma_n<\infty$. The inclusion is strict as if we choose the sequences $q=\{\gamma_1^{-\frac{1}{p}}, \frac{\gamma_1^{-\frac{1}{p}}}{2}, 0, \ldots\}$, and $\{\gamma_n\}\in l_1$ then $x=\{(-1)^{n+1}n\}\in \Gamma_p$ but $x\notin l_\infty$.
\end{proof}
In the next result, we obtain the associate space of $\Gamma_p$ for $p>1$. The associate space (known as $\beta$-dual or K\"{o}the-Toeplitz dual in literature) of $X$ (see \cite{MADDOXLN}) is denoted by $X^{'}$ or $X^{\beta}$, and defined as the space of all sequences $y=\{b_n\}_{n\geq 1}$ such that for every $x=\{a_n\}_{n\geq 1}\in X$, the series $\displaystyle\sum_{n=1}^{\infty}a_nb_n$ converges.
\begin{thm}
The associate space of $\Gamma_p$ is the set of all sequences $y=\{b_n\}_{n\geq 1}$ such that $\displaystyle \sup_{n\geq 1}\Big\{\sum_{k=1}^{n-1}|\gamma_k^{-\frac{1}{p}}(b_k-b_{k+1})|^q+ |\gamma_n^{-\frac{1}{p}}b_n|^q \Big\}^{\frac{1}{q}}<\infty$.
\end{thm}
\begin{proof}
Let $x=\{a_n\}_{n\geq 1}\in \Gamma_p$, and $y=\{b_n\}_{n\geq 1}\in (\Gamma_p)'$ be two sequences. Using the idea of Abel transformation, one can write
\begin{align}{\label{DUALI}}
\displaystyle\sum_{k=1}^{n}a_kb_k
& = \sum_{k=1}^{n-1}\Big(\gamma_k^{\frac{1}{p}}\sum_{i=1}^{k}q_ia_i\Big)\Big(\gamma_k^{-\frac{1}{p}}(b_k-b_{k+1})\Big) + \Big(\gamma_n^{\frac{1}{p}}\sum_{i=1}^{n}q_ia_i\Big)\Big(\gamma_n^{-\frac{1}{p}}b_n\Big) = \displaystyle\sum_{k=1}^{n} h_{nk}\sigma_k,
\end{align}where $\sigma_k=\gamma_k^{\frac{1}{p}}\displaystyle\sum_{i=1}^{k}q_ia_i$, and the matrix $H=(h_{nk})_{n, k \geq 1}$ is given as below:
\begin{align*}
h_{nk} &= \left\{
\begin{array}{lll}
    \gamma_k^{-\frac{1}{p}}(b_k-b_{k+1}) & \quad 1\leq k \leq n-1,\\ \vspace{0.2cm}
    \gamma_n^{-\frac{1}{p}}b_n & \quad k=n\\
    0 & \quad k>n.
\end{array}\right.
\end{align*}
Note that for $x\in \Gamma_p$, the sequence $\sigma=(\sigma_k)_{k\geq 1}$ is in $l_p$. Since for each $k\in \mathbb{N}$, the sequence $\{h_{nk}\}_{n=1}^{\infty}$ converges so $H\sigma\in c$, where $c$ is the space of all convergent sequences. Therefore, from (\ref{DUALI}), we conclude that for $x\in \Gamma_p$, sequence $\{a_nb_n\}_{n\geq 1}\in cs$, where $cs$ is the space of all convergent series if and only if $H\sigma\in c$ whenever $\sigma\in l_p$. This yields the fact that $y\in (\Gamma_p)'$ iff $H: l_p\rightarrow c$. It is known that from Stieglitz and Tietz result \cite{STTIZ} that the map $H: l_p\rightarrow c$ is possible if and only if the following conditions hold:\\
$(i)$ $\displaystyle\lim_{k\rightarrow\infty}h_{nk}$ exists for each $k\in \mathbb{N}$, and\\
$(ii)$ $\displaystyle\sup_{n\geq 1} \Big(\sum_{k=1}^{n}|h_{nk}|^q\Big)^{\frac{1}{q}}<\infty$, where $\frac{1}{p}+\frac{1}{q}=1$.\\
Since the first condition is trivially satisfied so the second condition gives the associate space $(\Gamma_p)'$. Hence the theorem is proved.
\end{proof}
For further results on the space $\Gamma_p$, we need to define a sequence $\{u_n\}_{n\geq 1}$ of sequences, where\\
$u_{1}=(\frac{\gamma_{1}^{-\frac{1}{p}}}{q_{1}}, -\frac{\gamma_{1}^{-\frac{1}{p}}}{q_{2}}, 0, 0,\cdots)$, \ldots,
$u_{n}=(0,\cdots 0, \frac{\gamma_{n}^{-\frac{1}{p}}}{q_{n}}, -\frac{\gamma_{n}^{-\frac{1}{p}}}{q_{n+1}}, 0, 0,\cdots)$, \ldots.\\
It is easy to verify that $\|u_n\|_{\Gamma_p}=1$ for each $n\in \mathbb{N}$. Then we have the following result.
\begin{thm}Let $x\in \Gamma_p$ for $p>1$. Then\\
(i) for every $x=\{a_n\}_{n\geq 1}$ we have $x=\displaystyle\sum_{i=1}^{\infty}\Big(\gamma_{i}^{\frac{1}{p}}\displaystyle\sum_{k=1}^{i}q_{k}a_{k}\Big)u_{i}$.\\
(ii) $\Gamma_p$ is separable.
\end{thm}
\begin{proof}
(i) It is sufficient to show that the sequence $\Big\{\displaystyle\sum_{i=1}^{n}\Big(\gamma_{i}^{\frac{1}{p}}\displaystyle\sum_{k=1}^{i}q_{k}a_{k}\Big)u_{i}\Big\}_{n\geq 1}$ norm convergent to $x$ for large $n$, and which can be easily deduce from the following computation.
\begin{align*}
\Big\|x-\displaystyle\sum_{i=1}^{n}\Big(\gamma_{i}^{\frac{1}{p}}\displaystyle\sum_{k=1}^{i}q_{k}a_{k}\Big)u_{i}\Big\|_{\Gamma_p}^p &= \Big\|\{0, \ldots, 0, a_{n+1}+\frac{1}{q_{n+1}}\displaystyle\sum_{k=1}^{n}q_{k}a_{k}, a_{n+2}, \ldots \}\Big\|_{\Gamma_p}^p\\
& = \displaystyle\sum_{k=n+1}^{\infty}\gamma_{k} \Big|\displaystyle\sum_{i=1}^{k}q_ia_{i}\Big|^{p}\rightarrow 0 \mbox{~as~} n\rightarrow\infty.
\end{align*}
(ii) We prove that there exists a countable dense subset of $\Gamma_p$. For each $n=1, 2, \ldots$, and fixed sequences $\{q_n\}$, $\{\gamma_n\}$, let us define a set $$S=\Big\{\displaystyle\sum_{k=1}^{n}\mu_k\Big(q_k\sum_{i=k}^{n}\gamma_{i}^{\frac{1}{p}}u_i\Big)=\Big\{\mu_1, \mu_2, \ldots, \mu_n, -\frac{1}{q_{n+1}}\displaystyle\sum_{k=1}^{n}q_{k}\mu_{k}, 0, 0, \ldots\Big\}: ~~\mu_k\in \mathbb{Q} \Big\}.$$ It is immediate that $S$ is countable and a subset of $\Gamma_p$. Also an element $\mu\in S$ can be written as $\mu=\displaystyle\sum_{i=1}^{n}\Big(\gamma_{i}^{\frac{1}{p}}\displaystyle\sum_{k=1}^{i}q_{k}\mu_{k}\Big)u_{i}$. Let $x\in \Gamma_p$ be any element. Since $\overline{\mathbb{Q}}=\mathbb{R}$, so one can have the following:
\begin{align*}
\big\|x-\mu\big\|_{\Gamma_p}^p &= \Big\|\Big\{a_1-\mu_1, a_2-\mu_2, \ldots, a_n-\mu_n, a_{n+1}+\frac{1}{q_{n+1}}\displaystyle\sum_{k=1}^{n}q_{k}a_{k}, a_{n+2}, \ldots \Big\}\Big\|_{\Gamma_p}^p\\
& = \displaystyle\sum_{k=1}^{n}\gamma_{k}q_k|(a_k-\mu_k)|^p+\displaystyle\sum_{k=n+1}^{\infty}\gamma_{k} \Big|\displaystyle\sum_{i=1}^{k}q_ia_{i}\Big|^{p}\rightarrow 0 \mbox{~as~} n\rightarrow\infty.
\end{align*}
Hence $\Gamma_p$ is separable.
\end{proof}

\NI\textit{Acknowledgement:}
The authors are very much thankful to Prof. David Krej\v{c}i\v{r}\'{i}k (Prague, Czech Republic) for his valuable comments on the first draft of our manuscript.

redpajama_set_name: RedPajamaArXiv


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text:
Meet the Jackfish: Matt Hilderbrandt
Matt Hilderbrandt looks more like an offensive lineman than a first baseman.
The 24-year-old Lindsay native is listed at 6-foot-4, 275 pounds and appears tough as nails.
But looks can be deceiving.
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Hilderbrandt make an immediate impact with the Jackfish last season batting a robust .347 with seven homers and 33 runs batted in and was named to the Intercounty Baseball League's second all-star team.
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"Ess reached out and told me all about it and it sounded awesome and it's even better than he said," Hilderbrandt said. "He said fans are great and a lot of people come out and support it and then I get here and he undersold it. It's unbelievable how many come out and support it and how everyone loves it here."
Hilderbrandt played both hockey and baseball growing up — his father was drafted by the Peterborough Petes — but ultimately hung up his skates in favour of cleats.
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He enjoyed his experience on the diamond south of the border.
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While at Limestone, Hilderbrandt got into body building at the behest of a teammate.
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Hilderbrand, who graduated with a major in liberal studies with a minor business and psychology, is working with his father at a tire store in his hometown.
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Zucco mines bronze at OFSAA
Roma's man in the middle
Meet the Jackfish: Jake Harford
Meet the Jackfish: Mattingly Romanin
Meet the Jackfish: Justin Gideon
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text:
Lepidosaphes lidgetti är en insektsart som först beskrevs av Cockerell 1899.  Lepidosaphes lidgetti ingår i släktet Lepidosaphes och familjen pansarsköldlöss. Inga underarter finns listade i Catalogue of Life.

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redpajama_set_name: RedPajamaWikipedia


The College considers an academic infraction a very serious matter. Procedures and standards exist for reporting, investigating, and adjudicating alleged instances of academic infraction.
These procedures and standards are maintained by the Academic Infractions Board (AIB), which consists of students and faculty members. The AIB is a subcommittee of the faculty Committee on Academic Standards.
Instructors should respond to inquiries concerning the forms that academic dishonesty may take in the particular kinds of work required in their courses. Also, instructors are responsible for detecting instances of academic dishonesty and dealing with suspected instances according to the procedures adopted by the faculty. Those procedures are designed to make the responsibility of judging and penalizing instances of suspected academic dishonesty a collegiate matter.
Regularized procedures and standards exist for reporting, investigating, and adjudicating alleged instances of academic dishonesty. These procedures and standards are maintained by the Academic Infractions Board (AIB).
The AIB consists of six faculty members (serving two-year terms) and at least four student members. For each AIB hearing, at least three faculty members and two student members are needed to convene a hearing, unless an alternative arrangement is authorized by the Associate Provost overseeing the AIB. The faculty members of the AIB are elected by the faculty during the elections for faculty committees in the spring. At least two of the AIB faculty members must be tenured faculty. After the elections are held, the Provost appoints a chair and a vice chair to the AIB. This arrangement aims to distribute the Board's workload so that, whenever possible, the chair and the vice chair will alternate presiding AIB hearings. The student members are appointed, from among the members of the Student Council's Academic Affairs Committee, by September 1.
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A student accused of a Tier-1 case may not drop the course in question while the charges are pending. The student may elect to drop the course after the conclusion of the Tier-1 case as long as the student receives permission from all of the following individuals: the course instructor, the department or program chair, the Dean of Academic Advising and the Associate Provost overseeing the AIB.
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