Calculate the following limit: $$\lim\limits_{n \to\infty} \left(\frac{1}{\sqrt{n^2+1}} + \frac{1}{\sqrt{n^2+2}} + ... + \frac{1}{\sqrt{n^2+n}} \right)$$
I think this limit equals $1$. I am not sure. Tried using the squeeze theorem:
$$1 \leq \left(\frac{1}{\sqrt{n^2+1}} + \frac{1}{\sqrt{n^2+2}} + ... + \frac{1}{\sqrt{n^2+n}} \right) \leq n\cdot \frac{1}{\sqrt{n^2}}$$.
It's quite clear why the right hand side is bigger than the middle term, but is $1$ really smaller of equals to the middle term?
Please note I can't use integrals here nor taylor series.
Thanks!