real analysis - 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!

Answer

For every $n>0$,
$$\frac{n}{\sqrt{n^2+n}}\le\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+\cdots+\frac{1}{\sqrt{n^2+n}}\le\frac{n}{\sqrt{n^2+1}}$$

Can you continue with Squeeze theorem?