TITLE: 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 [7 upvotes]: 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.

REPLY [2 votes]: 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 $$\sum_{k=1}^{n^2+n} \frac1{\sqrt{k}} -\sum_{k=1}^{n^2} \frac1{\sqrt{k}}=2\sqrt{n^2+n}-2n+o(1)=2n\left(\sqrt{1+1/n}-1\right)+o(1)\to1.$$