TITLE: How to prove $(\frac{1}{5^3}-\frac{1}{7^3})+(\frac{1}{11^3}-\frac{1}{13^3})+(\frac{1}{17^3}-\frac{1}{19^3})+...=(1-\frac{\pi ^3}{18\sqrt{3}})$
QUESTION [11 upvotes]: How to prove 
$$ \sum_{k=1}^\infty \left[\frac{1}{(6k-1)^3} - \frac{1}{(6k+1)^3}\right] = 1 - \frac{\pi^3}{18\sqrt{3}}$$
I think this equality likes the Dirichlet Beta function.  The numerical value is checked but I don't have the proving.  Any help

REPLY [6 votes]: Approach, which uses Fourier series.
Denote 
$$
S = \sum_{k=1}^\infty \left[\frac{1}{(6k-1)^3} - \frac{1}{(6k+1)^3}\right].
$$
Consider function
$$
f(x) = \dfrac{\pi x(\pi-x)}{8}, \qquad x\in[0,\pi];\tag{1}
$$
construct the odd extension of $f(x)$ to the interval $[−\pi, \pi]$:
$f(-x)=-f(x), x\in [0,\pi]$;
and make it $2\pi$-periodic: copy to each segment $[\pi (2k-1), \pi(2k+1)]$, $k\in\mathbb{Z}$.
Then, function $f(x)$ is odd, $2\pi$-periodic, and it belongs to class $C^1$.
Fourier series of this function:
$$
f(x) = \sum_{k=1}^{\infty} \dfrac{\sin(2k-1)x}{(2k-1)^3}= \dfrac{\sin x}{1^3}+\dfrac{\sin 3x}{3^3}+\dfrac{\sin{5x}}{5^3}+\dfrac{\sin{7x}}{7^3}+ \ldots .\tag{2}
$$
Consider $f(2\pi/3)$:
$(1)\Rightarrow$
$$
f(2\pi/3) = \dfrac{\pi^3}{36}.\tag{3}
$$
$(2)\Rightarrow$
$$
 f(2\pi/3) = \dfrac{\sin{2\pi/3}}{1^3}+\dfrac{\sin{6\pi/3}}{3^3}+\dfrac{\sin{10\pi/3}}{5^3}+\dfrac{\sin{14\pi/3}}{7^3}+\ldots \\
=\dfrac{\sqrt{3}}{2}\left(
\dfrac{1}{1^3}+\dfrac{0}{3^3}+\dfrac{-1}{5^3}+\dfrac{1}{7^3}+\dfrac{0}{9^3}+\dfrac{-1}{11^3}+\dfrac{1}{13^3}+\ldots
\right) = \dfrac{\sqrt{3}}{2}(1-S).\tag{4}
$$
$(3),(4) \Rightarrow $
$$
\dfrac{\pi^3}{36} = \dfrac{\sqrt{3}}{2}(1-S), 
$$
$$
1-S = \dfrac{\pi^3}{18\sqrt{3}}, 
$$
$$
S=1- \dfrac{\pi^3}{18\sqrt{3}}.
$$