Sum of the series $\sum\limits_{n=0}^\infty \frac{1}{(3n+1)^3}$ The following result matches very good numerically: 
$$\sum_{n=0}^\infty \frac{1}{(3n+1)^3}=\frac{13}{27}\zeta(3)+\frac{2\pi^3}{81\sqrt{3}}.$$
Though I'm not sure how to approach this. How can we prove it? Also, is it possible to find closed form for $$\sum_{n=0}^\infty \frac{1}{(3n+1)^5}$$ or the like?

Let $$S_1=\sum_{n=0}^{\infty} \frac{1}{(3n+1)^3} ,S_2=\sum_{n=0}^{\infty} \frac{1}{(3n+2)^3}.$$
It's easily seen that $$S_1+S_2+\sum_{n=1}^{\infty}\frac{1}{(3n)^3}=\zeta(3).$$
That is, $$S_1+S_2=\zeta(3)-\frac{1}{27}\zeta(3)=\frac{26}{27}\zeta(3)\tag{1}$$
However, $$S_1-S_2=\sum_{n=0}^{\infty} \frac{1}{(3n+1)^3}-\frac{1}{(3n+2)^3}=\sum_{n=-\infty}^{\infty} \frac{1}{(3n+1)^3}=\frac{4\pi^3}{81\sqrt{3}}\tag{2}$$
To calculate $(2)$, consider the Weierstrass form of the sinc function $$\frac{\sin x}{x}=\prod\limits_{n=1}^\infty\left(1-\frac{x^2}{\pi^2 n^2}\right)$$
and take $\frac{d^3}{dx^3}\log(\cdot)$ of both sides.
Set $x=\frac{\pi}{3}.$

Update.  I found that this question has been already asked here, so I link these two:        Sum related to zeta function
Also, I give an alternative way to calculate $(2)$. 
Notice that
$$\frac{2}{\sqrt{3}} \,\sin\left( \frac{2 \pi n}{3} \right) =\begin{cases} 0 & n\equiv 0 \\ 1 & n\equiv 1 \\ -1 & n\equiv 2
\end{cases} \pmod{3}.$$
Therefore $$S_1 - S_2 =\frac{2}{\sqrt{3}}\sum_{n=1}^{\infty} \frac{\sin(2\pi n/3)}{n^3}. \tag{3}$$
Finally, $(2)$ follows from the Fourier expansion $$ \sum_{n=1}^{\infty} \frac{\sin(2 \pi x n)}{n^3}=\frac{\pi^3}{3}\left( \{x\}-3\{x\}^2+2\{x\}^3\right), \tag{4}$$
where $\,\{x\}$ is the fractional part of $x$.
