# How to prove that $\lim_{n\to\infty}\left( \sum_{r=1}^n \dfrac 1 {\sqrt{n^2 + r}} \right) = 1$

We have to show that
$$\lim_{n\to\infty} \left( \dfrac 1 {\sqrt{n^2 +1}} +\dfrac 1 {\sqrt{n^2 +2}} + \dfrac 1 {\sqrt{n^2 +3}} + \ldots + \dfrac 1 {\sqrt{n^2 +n}} \right) = 1$$

Now, I thought of doing it as a definite integral, however the terms don’t translate into a function of $\left(\dfrac r n\right)$. I tried searching wolfram-alpha, but it used Hurwitz-zeta function, of which I have no clue. (This is an assignment problem)

#### Solutions Collecting From Web of "How to prove that $\lim_{n\to\infty}\left( \sum_{r=1}^n \dfrac 1 {\sqrt{n^2 + r}} \right) = 1$"

Hint: For all $n\in \mathbb N$ it holds that

$$\dfrac n {\sqrt{n^2 +n}} \leq \dfrac 1 {\sqrt{n^2 +1}} +\dfrac 1 {\sqrt{n^2 +2}} + \dfrac 1 {\sqrt{n^2 +3}} + \ldots + \dfrac 1 {\sqrt{n^2 +n}}\leq \dfrac n {\sqrt{n^2 +1}}.$$