Computing a series rised from definite integral of a floor function

Question

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.

Answer 1

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