How to prove $(\frac{1}{5^3}-\frac{1}{7^3})+(\frac{1}{11^3}-\frac{1}{13^3})+(\frac{1}{17^3}-\frac{1}{19^3})+...=(1-\frac{\pi ^3}{18\sqrt{3}})$ 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
 A: 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 $
$$
\dfrac{\pi^3}{36} = \dfrac{\sqrt{3}}{2}(1-S), 
$$
$$
1-S = \dfrac{\pi^3}{18\sqrt{3}}, 
$$
$$
S=1- \dfrac{\pi^3}{18\sqrt{3}}.
$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$\ds{\sum_{k\ =\ 1}^{\infty}\bracks{
     {1 \over \pars{6k - 1}^{3}} - {1 \over \pars{6k + 1}^{3}}}
      =1 - {\pi^{3} \over 18\root{3}}:\ {\large ?}}$.

\begin{align}&\color{#66f}{\large
\sum_{k\ =\ 1}^{\infty}\bracks{
{1 \over \pars{6k - 1}^{3}} - {1 \over \pars{6k + 1}^{3}}}}
=\left.\half\,\partiald[2]{}{\mu}
\sum_{k\ =\ 1}^{\infty}\pars{
{1 \over 6k - \mu} - {1 \over 6k + \mu}}\right\vert_{\,\mu\ =\ 1}
\\[5mm]&=\left. {1 \over 12}\,\partiald[2]{}{\mu}
\sum_{k\ =\ 0}^{\infty}\pars{
{1 \over k + 1 - \mu/6} - {1 \over k + 1 + \mu/6}}\right\vert_{\,\mu\ =\ 1}
\\[5mm]&=\left. {1 \over 12}\,\partiald[2]{}{\mu}
\bracks{\Psi\pars{1 + {\mu \over 6}} - \Psi\pars{1 - {\mu \over 6}}}
\right\vert_{\,\mu\ =\ 1}
={1 \over 432}\bracks{\Psi''\pars{7 \over 6} - \Psi''\pars{5 \over 6}}
\\[5mm]&={1 \over 432}\bracks{
\Psi''\pars{1 \over 6} + {2 \over \pars{1/6}^{3}}- \Psi''\pars{5 \over 6}}
={1 \over 432}\bracks{432 + 
\pi\,\totald[2]{\cot\pars{\pi\mu}}{\mu}}_{\mu\ =\ 5/6}
\\[5mm]&=1 + {1 \over 432}\,\bracks{2\pi^{3}\cot\pars{5\pi \over 6}
\csc^{2}\pars{5\pi \over 6}}
=\color{#66f}{\large 1 - {\pi^{3} \over 18\root{3}}}
\end{align}


$\ds{\Psi}$ is the Digamma Function and we used the identities:

\begin{align}
\sum_{k\ =\ 0}{1 \over \pars{k + x}\pars{k + y}}
&={\Psi\pars{x} - \Psi\pars{y} \over x - y}
\\[5mm]\Psi''\pars{1 + z}&=\Psi''\pars{z} + {2 \over z^{3}}
\\[5mm]
\Psi''\pars{1 - z}&=\Psi''\pars{z} + \pi\,\totald[2]{\cot\pars{\pi z}}{z}
=\Psi''\pars{z} + 2\pi^{3}\cot\pars{\pi z}\csc^{2}\pars{\pi z}
\end{align}
A: Notice$\color{blue}{^{[1]}}$
$$\sum_{k=1}^\infty \left( \frac{1}{(6k-1)^3} - \frac{1}{(6k+1)^3}\right)
= \sum_{\substack{k=-\infty\\ k\ne 0}}^\infty \frac{1}{(6k-1)^3}
= 1 - \frac{1}{6^3}\sum_{k=-\infty}^\infty \frac{1}{(\frac16-k)^3}$$
Recall the infinite product expansion of $\sin x$
$$\sin x = x \prod_{k=1}^\infty \left( 1 - \frac{x^2}{k^2\pi^2}\right)$$
If one take logarithm and differentiate, one obtain an expansion of $\cot x$
$$\cot x = \sum_{k=-\infty}^\infty \frac{1}{x - k\pi}
\quad\iff\quad
\sum_{k=-\infty}^\infty \frac{1}{x-k} = \pi\cot(\pi x)\tag{*1}
$$
Differentiate the expansion on the right two more times, we get$\color{blue}{^{[2]}}$
$$\sum_{k=-\infty}^\infty \frac{1}{(x-k)^3} 
= \frac12 \frac{d^2}{dx^2} \left[ \sum_{k=-\infty}^\infty \frac{1}{x-k} \right]
= \frac{\pi}{2} \left[ \frac{d^2}{dx^2}\cot(\pi x) \right]
= \frac{\pi^3 \cos(\pi x)}{\sin(\pi x)^3}$$
As a result, the sum we want is
$$\sum_{k=1}^\infty \left( \frac{1}{(6k-1)^3} - \frac{1}{(6k+1)^3}\right)
= 1 - \frac{\pi^3}{6^3}\frac{\cos\frac{\pi}{6}}{\sin(\frac{\pi}{6})^3} \
= 1 - \frac{\pi^3}{18\sqrt{3}}
$$
Notes


*

*$\color{blue}{[1]}$ - Infinite sum of the form $\sum\limits_{k=-\infty}^\infty (\cdots)$ should be interpreted as $\lim\limits_{N\to\infty}\sum\limits_{k=-N}^N (\cdots)$.

*$\color{blue}{[2]}$ - To those who are not comfortable with the use of differentiation of 
an expansion. An alternate approach is start from the more well known expansion $(*1)$,
compute a contour integral of the form:
$$\frac{1}{2\pi i}\int_{|z|=R} \frac{\pi\cot(\pi z)}{(\frac16 - z)^3} dz$$
and show it vanishes as $R \to \infty$. 
The sum of the residues from $z \in \mathbb{Z}$  will be equal to the sum $\sum\limits_{k=-\infty}^\infty \frac{1}{\left(\frac16 - k\right)^3}$. It will be compensate by the residue
of the pole at $z = \frac16$. Since the pole at $z = \frac16$ is a triple one, its contribution will be proportional to $\frac{d^2}{dx^2}\pi\cot(\pi x)$.

