Using Fourier Series to Find Infinite Sum I'm trying to use the Fourier series for $f(x)=x^3$ on $[-\pi,\pi]$ to show that 
$$\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{(2k-1)^3}=\frac{\pi^3}{32} $$
I've found the Fourier series to be
$$S(f)(x)= 2\sum_{k=1}^{\infty} \frac{(-1)^k6+(-1)^{k+1}\pi^2k^2}{k^3} \sin(kx)$$
I've tried evaluating at $x=\frac{\pi}{2}$ to get the correct result, but the my equation does not simplify correctly. The even $k$ terms vanish because of the $\sin(kx)$, but I can't seem to get rid of the "$\pi^2$" on the righthand side.
Suggestions or hints would be nice. Thanks!
 A: Use the Fourier series of $x^3 - \pi^2 x$ instead.
A: When $x=\pi/2$, the series generates 
$$
f(\pi/2) = \frac{\pi^3}{8}=-2\sum_{k=1}^{\infty}(-1)^k\frac{\pi^2k^2-6}{k^3}\sin(k\pi/2) = 2(\pi^2 - 6) + 0 -\frac{2}{27}(9\pi^2-6)+0+\frac{2}{125}(25\pi^2-6)+0-\cdots\\
$$
If the Fourier series converges absolutely at one $x$, then it converges absolutely for all $x$. Since $f$ is a Fourier series, by Parseval's theorem, there exist a function $g$ such that $\int\lvert g(x)\rvert^2dx = 2\pi\sum\lvert a_k\rvert < \infty$. Thus, the series converges absolutely for all $x$; therefore, we can rearrange the terms.
\begin{align}
\frac{\pi^3}{8} &= -12\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{(2k-1)^3}+2\pi^2\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{2k-1}\\
\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{(2k-1)^3}&=\frac{\pi^2}{6}\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{2k-1}-\frac{\pi^3}{96}\tag{1}
\end{align}
I trust you can solve the series in equation $(1)$.
A: Another helpful strategy for this problem would be to look at ways to re-index your series to match up with the given series. Right now your bottom term contains $k^3$ rather than the $(2k-1)^3$ of the desired solution.  The most straightforward way would be to re-index the even and odd terms (like how a series expansion of $e^{ix}=\cos(x)+i\sin(x)$ works) as shown below with the $\sin(kx)$ shown explicitly:
$$
\sum_{k=1}^{\infty} f_k(x) \cdot \sin(kx)=
\sum_{k'=1}^{\infty} \bigg[f_{2k'}(x) \cdot  \sin(2k'x) + f_{2k'-1}(x) \cdot  \sin((2k'-1)x)\bigg]
$$
This is helpful in showing how your "even k terms vanish" since $\sin(2k'x)\to\sin(k'\pi)$ when you set $x=\pi/2$ while $\sin((2k'-1)x)\to-\cos(k'\pi)=(-1)^{k'+1}$. Then the last steps would be to show how what remains can be simplified (using $\pi/4$) as others have already described in your question's comments.
