Can someone explain the limit $\lim _{n \rightarrow \infty} \left(\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+\cdots+\frac{1}{\sqrt{n^2+n}}\right)$?

Question

$$\begin{aligned} &\text { Find the following limit: } \lim _{n \rightarrow \infty} \left(\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+\cdots+\frac{1}{\sqrt{n^2+n}}\right)\\ &\text { **In a solution I got** Ans:- Let } u_n=\frac{n}{\sqrt{n^2+n}}\\ &\therefore \lim _{n \rightarrow \infty} u_n=\lim _{n \rightarrow \infty} \frac{n}{\sqrt{n^{2}+n}}=\lim _{n \rightarrow \infty} \frac{1}{\sqrt{1+\frac{1}{n}}}=1\\ &\text { By Cauchy's first theorem:- } \lim _{n \rightarrow \infty} \left(\frac{u_1+\cdots+u_n}{n}\right)=1\\ &\text{So, } \lim _{n \rightarrow \infty}\left(\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+\cdots+\frac{1}{\sqrt{n^2+n}}\right)=1 \end{aligned}$$

I am unable to understand this. How the general term comes like this! Any such explanation to the whole problem would be greatly appreciated.

Answer 1

$$\frac{1}{\sqrt{n^2+n}} \leq\frac{1}{\sqrt{n^2+1}}\leq \frac{1}{\sqrt{n^2}}\\ \frac{1}{\sqrt{n^2+n}} \leq\frac{1}{\sqrt{n^2+2}}\leq \frac{1}{\sqrt{n^2}}\\ \frac{1}{\sqrt{n^2+n}} \leq\frac{1}{\sqrt{n^2+3}}\leq \frac{1}{\sqrt{n^2}}\\ \vdots\\ \frac{1}{\sqrt{n^2+n}} \leq\frac{1}{\sqrt{n^2+n}}\leq \frac{1}{\sqrt{n^2}}.$$ 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.

Answer 2

You cannot apply Cauchy's first theorem because $$\sum_{k=1}^n\frac{1}{\sqrt{n^2+k}}\ne \frac{\sum_{k=1}^nu_k}{n}$$

since $$\sum_{k=1}^nu_k=\sum_{k=1}^n\frac{k}{\sqrt{k^2+k}}$$