text: # Find $\lim_{n\rightarrow \infty} \frac{1}{n^2} + \frac{2}{(n+1)^2}+\dots+\frac{n+1}{(2n)^2}$
  
  I want to find the following limit: $$\lim_{n\rightarrow \infty} \frac{1}{n^2} + \frac{2}{(n+1)^2}+ \frac{3}{(n+2)^2}+\dots+\frac{n+1}{(2n)^2}$$ I guess it converges to 0 and I tried to prove it using the squeeze theorem but it doesn't seem to work. Any ideas?
  
  • What do you mean by $n\rightarrow 0$? – Yujie Zha Jan 7 '18 at 23:24
  • my mistake, I edited it – asdf Jan 7 '18 at 23:29
  • I don’t know if it helps, but this can also be written as $$\lim_{n\to\infty}\sum_{x=0}^n\frac{x+1}{(x+n)^2}$$ Sometimes rewriting in sigma notation helps me – gen-z ready to perish Jan 8 '18 at 0:16
  
  Your limit can be rewritten as a Riemann sum + $a_n$ where $a_n\rightarrow 0$.
  
  This is a Riemann sum:
  
  \begin{align} \lim_{n\rightarrow \infty} \frac{1}{(n+1)^2}+ \frac{2}{(n+2)^2}+\dots+\frac{n}{(2n)^2}&=\lim_{n\rightarrow \infty} \frac{1}{n}\left ( \frac{1/n}{(1+1/n)^2}+ \frac{2/n}{(1+2/n)^2}+\dots+\frac{n/n}{(1+n/n)^2}\right)\\ &=\int_{0}^1 \frac{x}{(1+x)^2} \text{d}x \\ &=\int_{0}^1 \frac{x+1}{(1+x)^2} \text{d}x -\int_{0}^1 \frac{1}{(1+x)^2} \text{d}x\\ &=\int_{0}^1 \frac{1}{1+x} \text{d}x -\int_{0}^1 \frac{1}{(1+x)^2} \text{d}x \end{align}
  
  Now, \begin{align}a_n&=\frac{1}{n^2} + \frac{1}{(n+1)^2}+ \frac{1}{(n+2)^2}+\dots+\frac{1}{(2n)^2} \\ &\leq \frac{1}{n^2} + \frac{1}{n^2}+ \frac{1}{n^2}+\dots+\frac{1}{n^2}\\ &=(n+1)\frac{1}{n^2}\rightarrow 0 \end{align}
  
  • Thanks for the answer, although I was looking for something simpler as it adresses to students not familiar with Riemann sums, just basic properties of sequences and limits – asdf Jan 8 '18 at 16:05
  
  Note that
  
  $$\frac{1}{n^2} + \frac{2}{(n+1)^2}+ \frac{3}{(n+2)^2}+\dots+\frac{n+1}{(2n)^2}=\sum_{k=n}^{2n}\frac{k-n+1}{k^2} =\sum_{k=n}^{2n}\frac{1}{k}-(n-1)\sum_{k=n}^{2n}\frac{1}{k^2}\sim \log \left(\frac{2n}{n-1}\right)-\frac{n+1}{2n}\to\log 2-\frac12$$
  
  indeed for Harmonic series and Euler–Maclaurin formula
  
  $$\sum_{k=n}^{2n}\frac{1}{k}=\sum_{k=1}^{2n}\frac{1}{k}-\sum_{k=1}^{n-1}\frac{1}{k}\sim\log2n-\log (2n-1)=\log \left(\frac{2n}{n-1}\right)$$
  
  $$\sum_{k=n}^{2n}\frac{1}{k^2}=\sum_{k=1}^{2n}\frac{1}{k^2}-\sum_{k=1}^{n-1}\frac{1}{k^2}\sim -\frac1{2n}+\frac{1}{n-1}=\frac{n+1}{2n(n-1)}$$
explanation: The document contains a math question about finding the limit of a series. The user also provides some context and their attempts to solve the problem, including a guess that the limit converges to 0 and an attempt to use the squeeze theorem. The document also includes comments and answers from other users, suggesting alternative approaches and providing a detailed solution using Riemann sums and properties of harmonic series.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: # Find $\lim_{n\rightarrow \infty} \frac{1}{n^2} + \frac{2}{(n+1)^2}+\dots+\frac{n+1}{(2n)^2}$
  
  I want to find the following limit: $$\lim_{n\rightarrow \infty} \frac{1}{n^2} + \frac{2}{(n+1)^2}+ \frac{3}{(n+2)^2}+\dots+\frac{n+1}{(2n)^2}$$ I guess it converges to 0 and I tried to prove it using the squeeze theorem but it doesn't seem to work. Any ideas?
  
  • What do you mean by $n\rightarrow 0$? – Yujie Zha Jan 7 '18 at 23:24
  • my mistake, I edited it – asdf Jan 7 '18 at 23:29
  • I don’t know if it helps, but this can also be written as $$\lim_{n\to\infty}\sum_{x=0}^n\frac{x+1}{(x+n)^2}$$ Sometimes rewriting in sigma notation helps me – gen-z ready to perish Jan 8 '18 at 0:16
  
  Your limit can be rewritten as a Riemann sum + $a_n$ where $a_n\rightarrow 0$.
  
  This is a Riemann sum:
  
  \begin{align} \lim_{n\rightarrow \infty} \frac{1}{(n+1)^2}+ \frac{2}{(n+2)^2}+\dots+\frac{n}{(2n)^2}&=\lim_{n\rightarrow \infty} \frac{1}{n}\left ( \frac{1/n}{(1+1/n)^2}+ \frac{2/n}{(1+2/n)^2}+\dots+\frac{n/n}{(1+n/n)^2}\right)\\ &=\int_{0}^1 \frac{x}{(1+x)^2} \text{d}x \\ &=\int_{0}^1 \frac{x+1}{(1+x)^2} \text{d}x -\int_{0}^1 \frac{1}{(1+x)^2} \text{d}x\\ &=\int_{0}^1 \frac{1}{1+x} \text{d}x -\int_{0}^1 \frac{1}{(1+x)^2} \text{d}x \end{align}
  
  Now, \begin{align}a_n&=\frac{1}{n^2} + \frac{1}{(n+1)^2}+ \frac{1}{(n+2)^2}+\dots+\frac{1}{(2n)^2} \\ &\leq \frac{1}{n^2} + \frac{1}{n^2}+ \frac{1}{n^2}+\dots+\frac{1}{n^2}\\ &=(n+1)\frac{1}{n^2}\rightarrow 0 \end{align}
  
  • Thanks for the answer, although I was looking for something simpler as it adresses to students not familiar with Riemann sums, just basic properties of sequences and limits – asdf Jan 8 '18 at 16:05
  
  Note that
  
  $$\frac{1}{n^2} + \frac{2}{(n+1)^2}+ \frac{3}{(n+2)^2}+\dots+\frac{n+1}{(2n)^2}=\sum_{k=n}^{2n}\frac{k-n+1}{k^2} =\sum_{k=n}^{2n}\frac{1}{k}-(n-1)\sum_{k=n}^{2n}\frac{1}{k^2}\sim \log \left(\frac{2n}{n-1}\right)-\frac{n+1}{2n}\to\log 2-\frac12$$
  
  indeed for Harmonic series and Euler–Maclaurin formula
  
  $$\sum_{k=n}^{2n}\frac{1}{k}=\sum_{k=1}^{2n}\frac{1}{k}-\sum_{k=1}^{n-1}\frac{1}{k}\sim\log2n-\log (2n-1)=\log \left(\frac{2n}{n-1}\right)$$
  
  $$\sum_{k=n}^{2n}\frac{1}{k^2}=\sum_{k=1}^{2n}\frac{1}{k^2}-\sum_{k=1}^{n-1}\frac{1}{k^2}\sim -\frac1{2n}+\frac{1}{n-1}=\frac{n+1}{2n(n-1)}$$
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document contains a math question about finding the limit of a series. The user also provides some context and their attempts to solve the problem, including a guess that the limit converges to 0 and an attempt to use the squeeze theorem. The document also includes comments and answers from other users, suggesting alternative approaches and providing a detailed solution using Riemann sums and properties of harmonic series.
  
  Label: question-and-answer