Let $1 + \frac{1}{3^3} + \frac{1}{5^3} + \frac{1}{7^3} + \dots=s$, show that then $\sum_1^\infty\frac{1}{n^3}=\frac{8}{7}s$

Question

Let $1 + \frac{1}{3^3} + \frac{1}{5^3} + \frac{1}{7^3} + \dots=s$, show that then $\sum_1^\infty\frac{1}{n^3}=\frac{8}{7}s$.

This is the last part of a problem that I am working on. So far, we have shown that the cosine series for $x^2$ is

$x^2=\frac{\pi^2}{3}+4\sum_1^\infty (-1)^n\frac{\cos(nx)}{n^2}$

The sine series for $x^2$ is

$x^2=2\pi\sum_1^\infty(-1)^{n+1}\frac{\sin nx}{n}-\frac{8}{\pi}\sum_1^\infty\frac{\sin ((2n-1)x)}{(2n-1)^3}$

The sine series for x is

$x=2(\sin x -\frac{\sin 2x}{2}+ \frac{\sin 3x}{3}-\dots)$

and using the above we got that

$\frac{\pi^3}{32}=1 - \frac{1}{3^3}+\frac{1}{5^3}-\frac{1}{7^3}+\dots$.

Not sure if any of this matters, I just don't know how to use this information to solve the last part. I don't really know if this information is needed at all, but I assume it is.

Answer 1

\begin{align*} \sum_1^{\infty} \frac 1 {n^3} &= \sum_{\text{even}} \frac 1 {n^3} + s \\ &= \sum_1^{\infty} \frac{1}{(2k)^3} + s\\ &= \frac 1 8 \sum_1^{\infty} \frac 1 {k^3} + s \end{align*}

Rearrange for the desired result.

Answer 2

Let $\sum_{n=1}^\infty \frac{1}{(2n-1)^3}=s$, then we have \begin{eqnarray*} \sum_{n=1}^\infty \frac{1}{n^3} &=& \sum_{n=1}^\infty \sum_{m=0}^\infty \frac{1}{(2n-1)^3} \frac{1}{2^{3m}} \\ &=& \sum_{m=0}^\infty \frac{1}{8^m} \sum_{n=1}^\infty \frac{1}{(2n-1)^3} \\ &=& \frac{1}{1-\frac{1}{8}} s \\ &=& \frac{8}{7} s \end{eqnarray*} Where we first count all odd numbers, then the numbers divisble by $2$ but not $4$, then divisble by $4$ not $8$ et cetera. This gives us the sum over $m$ in the first line.
And we may switch order of summation as we sum only positive numbers.