How to prove $(\frac{1}{5^3}-\frac{1}{7^3})+(\frac{1}{11^3}-\frac{1}{13^3})+(\frac{1}{17^3}-\frac{1}{19^3})+...=(1-\frac{\pi ^3}{18\sqrt{3}})$

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 $\frac{d^2}{dx^2}\pi\cot(\pi x)$.

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 $

$$ \dfrac{\pi^3}{36} = \dfrac{\sqrt{3}}{2}(1-S), $$ $$ 1-S = \dfrac{\pi^3}{18\sqrt{3}}, $$ $$ S=1- \dfrac{\pi^3}{18\sqrt{3}}. $$