Proving the limit $\lim _{n\to \infty }\left(\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+...+\frac{1}{\sqrt{n^2+n}}\right)=1$

Question

I think I tried everything I know when trying to prove this limit $$\lim _{n\to \infty }\left(\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+...+\frac{1}{\sqrt{n^2+n}}\right)=1$$ but I think that either there is some rule that I don't know or I am simply missing the creativity needed to solve it.. Help would be very appreciated.

Answer 1

Hint: squeeze that thing

$$\frac n{\sqrt{n^2+n}}\le\frac1{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+...+\frac{1}{\sqrt{n^2+n}}\le\frac n{\sqrt{n^2+1}}$$

Answer 2

Very informally,

$$\lim _{n\to \infty }\frac1n\left(\frac{1}{\sqrt{1+\frac1{n^2}}}+\frac{1}{\sqrt{1+\frac2{n^2}}}+\cdots\frac{1}{\sqrt{1+\frac n{n^2}}}\right)=\lim _{n\to \infty }\int_0^1\frac{dx}{\sqrt{1+\frac xn}}.$$

Answer 3

You might consider the following "creative"...

It's easy to see that $a_n:=-2\sqrt{n} +\sum_{k=1}^n \frac1{\sqrt{k}}$ is decreasing, and since $$2\sqrt{n}-2=\int_1^n\frac{dx}{\sqrt{x}}\le\sum_{k=1}^n \frac1{\sqrt{k}}$$ we find $a_n$ converges. It thus suffices to note your expression can be written as $$\sum_{k=1}^{n^2+n} \frac1{\sqrt{k}} -\sum_{k=1}^{n^2} \frac1{\sqrt{k}}=2\sqrt{n^2+n}-2n+o(1)=2n\left(\sqrt{1+1/n}-1\right)+o(1)\to1.$$