Various evaluations of the series $\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^3}$

Method by Fourier Series

Consider the function $f(x) = x(1 - x)$, $0 \le x \le 1$. It has Fourier sine series expansion

$$f(x) = \frac{8}{\pi^3}\sum_{n = 1}^\infty \frac{1}{(2n-1)^3}\sin{(2n-1)\pi x}.$$

Setting $x = \frac{1}{2}$ results in

$$\frac{1}{4} = \frac{8}{\pi^3}\sum_{n = 1}^\infty \frac{(-1)^{n-1}}{(2n-1)^3},$$

or

$$\frac{\pi^3}{32} = \sum_{n = 1}^\infty \frac{(-1)^{n-1}}{(2n-1)^3}.$$

By reindexing the sum we can write

$$\frac{\pi^3}{32} = \sum_{n = 0}^\infty \frac{(-1)^n}{(2n+1)^3}.$$

Method by Contour Integration

Let $g(z) = \frac{1}{(2z - 1)^3}$. Then $g$ has only one pole of order $3$ at $z = \frac{1}{2}$. Let $N$ be a positive integer, and consider the contour integral

$$\frac{1}{2\pi i}\int_{\Gamma_N} \pi\csc \pi z\, g(z)\, dz,$$

where $\Gamma_N$ is a positively oriented square with vertices at $\left(N + \frac{1}{2}\right)(\pm 1 \pm i)$. The residue theorem gives

\begin{align}\frac{1}{2\pi i}\int_{\Gamma_N} \pi \csc \pi z\, g(z)\, dz &= \sum_{n = -N}^N \operatorname{Res}\limits_{z = n} \pi \csc \pi z\, g(z) + \operatorname{Res}\limits_{z = \frac{1}{2}} \pi \csc \pi z\, g(z)\\ &= \sum_{n = -N}^N (-1)^n g(n) + \frac{\pi^3}{16}. \end{align}

For $|z| \ge 1$, $|g(z)| \le |z|^{-3}$. Thus, $$\frac{1}{2\pi i}\int_{\Gamma_N} \pi \csc \pi z\, g(z)\, dz \to 0 \quad \text{as} \quad N \to \infty.$$

Hence

$$0 = \sum_{n = -\infty}^\infty (-1)^n g(n) + \frac{\pi^3}{16}$$

that is,

$$\frac{\pi^3}{16} = \sum_{n = -\infty}^\infty \frac{(-1)^{n-1}}{(2n-1)^3}.$$

Now

\begin{align}\sum_{n = -\infty}^\infty \frac{(-1)^{n-1}}{(2n-1)^3} &= \sum_{n = -\infty}^0 \frac{(-1)^{n-1}}{(2n-1)^3} + \sum_{n = 1}^\infty \frac{(-1)^{n-1}}{(2n-1)^3}\\ & = \sum_{n = 0}^\infty \frac{(-1)^{n}}{(2n+1)^3} + \sum_{n = 1}^\infty \frac{(-1)^{n-1}}{(2n-1)^3}\\ & = 2\sum_{n = 0}^\infty \frac{(-1)^n}{(2n+1)^3}. \end{align}

Thus

$$\frac{\pi^3}{16} = 2\sum_{n = 0}^\infty \frac{(-1)^n}{(2n+1)^3}.$$

Finally, we have

$$\frac{\pi^3}{32} = \sum_{n = 0}^\infty \frac{(-1)^n}{(2n+1)^3}.$$

Differentiating twice the logarithm of the Weierstrass representation of sine gives $$ \sum\limits_{n=-\infty}^{\infty} {1\over (z+n)^2}=\frac{\pi^{2}}{\sin^{2}(\pi z)} $$ (as i've been answered in here.)

Now differentiate once more and consider $z=\frac{1}{4}$.

I do not know how much this could help you; so forgive me if I am out off topic.

Rewriting a little the expression $$\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^3}x^n=\frac{1}{8} \,\Phi \left(-x,3,\frac{1}{2}\right)$$ where appears the Lerch transcendent function. Now, using $x=1$, we can get the result.