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.

$$ \sqrt{n+\frac{1}{n}} -\sqrt n = \frac{1}{n\left(\sqrt{n+\frac{1}{n}} +\sqrt n\right)} \leq \frac{1}{2n^{3/2}} .$$
Now
$$ \sum_{n=N+1}^\infty n^{-3/2} \leq \int_N^\infty x^{-3/2} dx =\frac{ 2}{\sqrt N}$$
