Given a function $$f(x)=x(\pi^2-x^2)$$
I calculated its Fourier series $$f(x)=x(\pi ^2-x^2)=\sum_{n=1}^{\infty} \frac{12(-1)^{n+1}\sin(nx)}{n^3}$$
Now i'm supposed to use this result to calculate the sum of $$\sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^3}$$
I tried to plug in a few values like $1/2 \pi or 3/2 \pi$ to make the $\sin$ go away but nothing worked out so far, how do i equate the denominator $(2n+1)^3$ and $(n)^3$
Thanks
Summation can be simplified as: $$ \frac{x}{12} (\pi^2 - x^2) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}\sin(nx)}{n^3} $$ for getting rid of $\sin(\theta)$. you can see $\theta=\frac{n\pi}{2}$ are special points. choose $x=\frac{\pi}{2}$ (as you correctly guessed!): $$ \sin (\frac{n\pi}{2}) = \begin{cases} 0 & n=2k \\ 1 & n=4k+1\\ -1 & n=4k+3\\ \end{cases} $$ so $\sin(nx)$ would be zero at even $n$. only nonzero elements are odd $n$'s. which in this case $(-1)^{n+1} =1$. so summation become: $$ \sum_{n=0}^{\infty} \frac{\sin((2n+1)\frac{\pi}{2})}{(2n+1)^3} $$ in which i just change $n$ to $2n+1$ because for even terms are zero and summation starts from $2n+1=1$, which in terms of $2n+1$ representation, n starts from $0$. $$ \sin ((2n+1)\frac{\pi}{2}) = \begin{cases} 1 & \text{n is even}\\ -1 & \text{n is odd}\\ \end{cases} = (-1)^n $$ so we get our series: $$ \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^3} = \frac{1}{12}f(\frac{\pi}{2}) =\frac{3\pi^3}{12\times8} = 0.9689461462593693. $$ which you can confirm numerically.