Can someone explain the limit $\lim _{n \rightarrow \infty} \left(\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+\cdots+\frac{1}{\sqrt{n^2+n}}\right)$? $$\begin{aligned}
&\text { Find the following limit: } \lim _{n \rightarrow \infty} \left(\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+\cdots+\frac{1}{\sqrt{n^2+n}}\right)\\
&\text { **In a solution I got** Ans:- Let } u_n=\frac{n}{\sqrt{n^2+n}}\\
&\therefore \lim _{n \rightarrow \infty} u_n=\lim _{n \rightarrow \infty} \frac{n}{\sqrt{n^{2}+n}}=\lim _{n \rightarrow \infty} \frac{1}{\sqrt{1+\frac{1}{n}}}=1\\
&\text { By Cauchy's first theorem:- } \lim _{n \rightarrow \infty} \left(\frac{u_1+\cdots+u_n}{n}\right)=1\\
&\text{So, } \lim _{n \rightarrow \infty}\left(\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+\cdots+\frac{1}{\sqrt{n^2+n}}\right)=1
\end{aligned}$$
I am unable to understand this. How the general term comes like this! Any such explanation to the whole problem would be greatly appreciated.

$$\frac{1}{\sqrt{n^2+n}} \leq\frac{1}{\sqrt{n^2+1}}\leq \frac{1}{\sqrt{n^2}}\\
\frac{1}{\sqrt{n^2+n}} \leq\frac{1}{\sqrt{n^2+2}}\leq \frac{1}{\sqrt{n^2}}\\
\frac{1}{\sqrt{n^2+n}} \leq\frac{1}{\sqrt{n^2+3}}\leq \frac{1}{\sqrt{n^2}}\\
\vdots\\
\frac{1}{\sqrt{n^2+n}} \leq\frac{1}{\sqrt{n^2+n}}\leq \frac{1}{\sqrt{n^2}}.$$ Now look at sum of them
$$\frac{n}{\sqrt{n^2+n}} \leq(\frac{1}{\sqrt{n^2+1}}+...+\frac{1}{\sqrt{n^2+n}})\leq \frac{n}{\sqrt{n^2}}.$$ Now take limit.
