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!
 A: 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
$$ \sum_{k=0}^{\infty} \frac{1}{(3k+1)^{3}} = \frac{13 \zeta(3)}{27} + \frac{2 \pi^{3}}{81 \sqrt{3}} .$$
A: $\newcommand{\+}{^{\dagger}}
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$\ds{}$
\begin{align}
\color{#66f}{\large\sum_{k = 0}^{\infty}{1 \over \pars{3k + 1}^{3}}}
&={1 \over 27}\sum_{k = 0}^{\infty}{1 \over \pars{k + 1/3}^{3}}
=\left. -\,{1 \over 27}\,\partiald{}{\mu}\sum_{k = 0}^{\infty}
{1 \over \pars{k + \mu}\pars{k + 1/3}}\,\right\vert_{\,\mu\ =\ {1/3}}
\\[3mm]&=-\,{1 \over 27}\,\partiald{}{\mu}\bracks{%
\Psi\pars{\mu} - \Psi\pars{1/3} \over \mu - 1/3}_{\mu\ =\ {1/3}}
=-\,{1 \over 54}\,\Psi''\pars{1 \over 3}
\\[3mm]&=\color{#66f}{\large{1 \over 243}\bracks{2\root{3}\pi^{3} + 117\zeta\pars{3}}} \approx 1.0208
\end{align}

See a Hurwitz Zeta Function link.

A: You can also start from $$\sum_{k=0}^{m} \frac{1}{(3k+1)^3}=\frac{1}{54} \left(\psi ^{(2)}\left(m+\frac{4}{3}\right)-\psi
   ^{(2)}\left(\frac{1}{3}\right)\right)$$ which simplifies to $$\sum_{k=0}^{m} \frac{1}{(3k+1)^3}=\frac{1}{54} \left(\psi ^{(2)}\left(m+\frac{4}{3}\right)+26 \zeta (3)+\frac{4 \pi
   ^3}{3 \sqrt{3}}\right)$$and take the limit for an infinite value of $m$. This leads to the answer given by Felix Marin and by Wolfram Alpha.
In fact, there is a nice generalization for $$\sum_{k=0}^{\infty} \frac{1}{(ak+b)^c}=a^{-c} \zeta \left(c,\frac{b}{a}\right)$$ in which appears Hurwitz Zeta function.
