A particular problem on series 
Problem: Show that there exist $c>0$ such that for all $N\in \mathbb N$ we have $$
\sum_{n=N+1}^{\infty}\left(\sqrt{n+\frac{1}{n}}-\sqrt{n}\right)\le \frac{c}{\sqrt{N}}
$$

I have no clue how to solve this. All I know is this fact $$\int_0^1\left(\sum_{n\in \mathbb N}\frac{1}{
\sqrt{n^3+nx}}\right)\mathrm dx=\sum_{n\in \mathbb N}\left(\int_0^1\frac{1}{
\sqrt{n^3+nx}}\mathrm dx \right)$$
$$=2\sum_{n=1}^{\infty}\left(\sqrt{n+\frac{1}{n}}-\sqrt{n}\right)$$
as the series of function is uniformly convergent .
Note: I am quoting this fact regarding the series of function because the above problem was meant to be solved as a consequence of this fact . But other methods are welcome as well.

A possible approach is to use Bernoulli's inequality:
$$
\sqrt{n+\frac{1}{n}} = \sqrt{n} \sqrt{1+\frac{1}{n^2}} \le \sqrt{n} \left( 1 + \frac{1}{2n^2}\right)
$$
That leaves you with
$$
 \sum_{n=N+1}^\infty \frac{1}{2n^{3/2}}\le \int_N^\infty \frac{1}{2x^{3/2}}\mathrm dx
$$
which  can be estimated above by the corresponding integral.
