Sum related to zeta function I was trying to evaluate the following sum: 
$$\sum_{k=0}^{\infty} \frac{1}{(3k+1)^3}$$
W|A gives a nice closed form but I have zero idea about the steps involved to evaluate the sum. How to approach such sums?
Following is the result given by W|A:
$$\frac{13\zeta(3)}{27}+\frac{2\pi^3}{81\sqrt{3}}$$
Any help is appreciated. Thanks!

Notice that
$$ \sum_{k=0}^{\infty} \frac{1}{(3k+1)^{3}} = \frac{1}{27} \sum_{n=0}^{\infty} \frac{1}{(k+\frac{1}{3})^{3}} = - \frac{1}{54} \psi_{2}\left(\frac{1}{3} \right) $$
where $\psi_{2}(x)$ is the second derivative of the digamma function.
Differentiating the  multiplication formula for the digamma function twice and letting $q=3$,
$$\psi_{2}(x) + \psi_{2} \left( x+ \frac{1}{3} \right) + \psi_{2} \left(x+ \frac{2}{3} \right) = 27 \psi_{2}(3x) .$$
Therefore, $$ \begin{align} \psi_{2} \left(\frac{1}{3} \right) + \psi_{2} \left( \frac{2}{3} \right) &=  27 \psi_{2}(1) - \psi_{2}(1) \\ &=27 \left( -2 \zeta(3) \right) + 2 \zeta(3) = -52 \zeta(3). \tag{1}\end{align} $$
And differentiating the reflection formula for the digamma function twice, 
$$ \psi_{2} (x) - \psi_{2}(1-x) = - 2\pi^{3} \cot(\pi z) \csc^{2}(\pi z) .$$
Therefore, $$\psi_{2} \left(\frac{1}{3} \right) - \psi_{2} \left( \frac{2}{3}\right) = - \frac{8 \pi^{3}}{3 \sqrt{3}} . \tag{2}$$
Adding $(1)$ and $(2)$,
$$ \psi_{2} \left( \frac{1}{3}\right) = -26 \zeta(3)  - \frac{4 \pi^{3}}{3 \sqrt{3}} .$$
So
$$ \sum_{k=0}^{\infty} \frac{1}{(3k+1)^{3}} = \frac{13 \zeta(3)}{27} + \frac{2 \pi^{3}}{81 \sqrt{3}} .$$
