How can you solve for the summation term, $\sum_{n=1}^{\infty} \frac{(-1)^n}{(n\pi)^2}$, in the Fourier series below?
$$ x^2 \thicksim \frac{1}{3} + 4\sum_{n=1}^{\infty} \frac{(-1)^n}{(n\pi)^2}\cos{n\pi x} $$
After rearranging the terms, I'm stuck at dividing out the cosine term from the summation.
$$ \frac{x^2}{4} - \frac{1}{12} \thicksim \sum_{n=1}^{\infty} \frac{(-1)^n}{(n\pi)^2}\cos{n\pi x} $$
Try plugging in $x=0$ to the original expression. Recall, $\cos(0)=1$.
Factor out the $$\frac {-1}{\pi^2}$$
To turn the question into $$\frac {-1}{\pi^2}\sum_{n=1}^{\infty}\frac {(-1)^{n+1}}{n^2}$$
Using this link we have $$\sum_{n=1}^{\infty}\frac {(-1)^{n+1}}{n^2}=\frac {\pi^2}{12}$$
Hence the answer is $\frac {-1}{12}$
First, note that $\sum1/n^{2}$ is convergent. Therefore, $$ \sum_{n\geq1}\frac{(-1)^{n}}{n^{2}}=\sum_{n\geq1}\frac{1}{(2n)^{2}}-\sum_{n\geq0}\frac{1}{(2n+1)^{2}}=\frac{1}{4}\left(\sum_{n\geq1}\frac{1}{n^{2}}-\sum_{n\geq0}\frac{1}{(n+1/2)^{2}}\right)=\frac{1}{4}\left(\zeta(2)-\zeta(2,1/2)\right). $$ $\zeta(2)=\pi^{2}/6$ is the solution of the Basel problem. As for the second term, you can use a well-known identity regarding the Hurwitz zeta function to get $\zeta(2,1/2)=(4-1)\zeta(2)=\pi^{2}/2$. Therefore, $$ \sum_{n\geq1}\frac{(-1)^{n}}{n^{2}}=-\frac{\pi^{2}}{12}. $$ Divide by $\pi^2$ to get $-1/12$.
