Calculate the limit $\left(\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+...+\frac{1}{\sqrt{n^2+n}}\right)$ I have to calculate the following limit:
$$\lim_{n\to\infty} \left(\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+...+\frac{1}{\sqrt{n^2+n}}\right)$$
I believe the limit equals $1$, and I think I can prove it with the squeeze theorem, but I don't really know how.
Any help is appreciated, I'd like to receive some hints if possible.
Thanks!

This is the final answer I got, thanks to all the help:
For every $n>0$,$\frac{n}{\sqrt{n^2+n}}\le \left(\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+...+\frac{1}{\sqrt{n^2+n}}\right) \le \frac{1}{\sqrt{n^2+n}}$
Using the squeeze theorem, we calculate the middle expression's limit.
$\lim_\limits{n \to \infty} \frac{n}{\sqrt{n^2+n}}=\lim_\limits{n \to \infty} \frac{\sqrt{n^2}}{\sqrt{n^2+n}}= \lim_\limits{n \to \infty}\sqrt{\frac{ {n^2}}{{n^2+n}}}=\lim_\limits{n \to \infty}\sqrt{\frac{ \frac{n^2}{n^2}}{{\frac{n^2}{n^2}+\frac{n}{n^2}}}}=\lim_\limits{n \to \infty} \sqrt{\frac{1}{1+\frac{1}{n}}}=1$ from limits arithmetic.
Likewise, we can calculate the right hand side and reach to the conclusion that the original sequence approaches $1$.
