Computing a series rised from definite integral of a floor function I interested to compute $$\int_0^1\left[\frac{1}{\sqrt{x}}\right]dx.$$
So, I did like this:
$$\int_0^1\left[\frac{1}{\sqrt{x}}\right]dx=\sum_{n=1}^\infty
\int_{1/(n+1)^2}^{1/n^2} ndx=\\
 \sum_{n=1}^{\infty}\frac{2n+1}{n(n+1)^2}=\\
=\sum_{n=1}^\infty  \frac{2}{(n+1)^2}+\sum_{n=1}^\infty \frac{1}{n(n+1)^2}=\frac{\pi^2}{3}-2+\sum_{n=1}^\infty \frac{1}{n(n+1)^2}$$
Now, how to compute the last series? Is it computible or it relates to $\zeta(3)$?
Thanks.

$\sum_{n=1}^\infty \frac{1}{n(n+1)^2}=\sum_{n=1}^\infty \frac{n+1-n}{n(n+1)^2}=\sum_{n=1}^\infty \frac{1}{n(n+1)}-\sum_{n=1}^\infty \frac{1}{(n+1)^2}=1-\sum_{n=1}^\infty \frac{1}{(n+1)^2}$
