TITLE: An arctan series with roots $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} \arctan \frac{1}{n \sqrt{n}}$
QUESTION [8 upvotes]: Solving an exercise today I came across this series and I'm curious to know if we can evaluate it. Here it is:
$$\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} \arctan \frac{1}{n \sqrt{n}}$$
It rings me bells about some other series with arctan's I have come across but I could not see any similarity on how to begin. Wolfram gives an approximation of $1.41379$. Note that $\sqrt{2} \approx 1.4142$. Too sad !!

REPLY [4 votes]: Let 
$$F(a) = \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} \arctan \frac{a}{n \sqrt{n}}$$
$$F'(a) = \sum_{n=1}^{\infty} \frac{n}{a^2+n^3}$$
Using wolfram we get 
$$F'(a) = -\frac{1}{3} \sum \frac{\psi(-\omega)}{\omega+1}$$
Which is summed over the roots of the equation $$\omega^3+3\omega^2+3\omega+1+a^2=0$$
I am not sure of the complexity of finding the anti derivative.