<s>[INST] Sum : $\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^3}$ Prove that : $$\sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2k+1)^3}=\frac{\pi^3}{32}.$$

I think this is known (see here), I appreciate any hint or link for the solution (or the full solution).
 [/INST] Here is another way using Fourier analysis:
Let
\begin{equation*}
  f(t)=\begin{cases}
    t-t^2 & 0<t<1 \\
    -f(-t) & -1 < t < 0
  \end{cases}
\end{equation*}
be a function with period 2. Then we can express $f$ in a Fourer series:
\begin{equation*}
  f(t)=\frac{a_0}{2}+\sum_{n=1}^{\infty}a_n \cos n\pi t+b_n \sin n\pi t
\end{equation*}
where
\begin{align*}
  &a_n=\frac{1}{1}\int_{0}^{2}f(t)\cos n\pi t \, dt \\
  &b_n=\frac{1}{1}\int_{0}^{2}f(t)\sin n\pi t\, dt
\end{align*}
But $f$ is odd, so $a_n=0$.
It follows that
\begin{align*}
  b_n&= \int_{0}^{2}f(t)\sin n\pi t\, dt=\{ f(t)\sin n\pi t \text{ even}\}=2\int_{0}^{1}f(t)\sin n\pi t\, dt = \\
  &= 2\int_{0}^{1}(t-t^2)\sin n\pi t\, dt=\frac{4-4(-1)^n}{n^3 \pi^3}
\end{align*}
Plugging in $t=\frac{1}{2}$, we get
\begin{equation*}
  \frac{1}{4}=4\sum_{n=1}^{\infty}\frac{1-(-1)^n}{\pi^3n^3}\sin \frac{n\pi}{2}
\end{equation*}
But $1-(-1)^n=0$ only if $n$ is even, so
\begin{equation*}
  \frac{1}{16}=\sum_{k=0}^{\infty}\frac{2(-1)^k}{(2k+1)^3\pi^3}\iff \sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)^3}=\frac{\pi^3}{32}
\end{equation*}
</s>