# Showing $\sum_{n=1}^\infty\frac{1-(-1)^n}{n^2} = \frac{\pi^2}{4}$ from Fourier expansion

Find the Fourier series for the function:

$$f(x) = \begin{cases} \hfill 0 \hfill & \text{- \pi<x< 0} \\ \hfill \pi - x \hfill & \text{ 0<x< \pi} \\ \end{cases}$$

on $- \pi < x< \pi$. Hence show that:

$$\sum_{n=1}^\infty\frac{1-(-1)^n}{n^2} = \frac{\pi^2}{4}$$

My attempt:

Fourier series general form:

$$FS\left[f(t)\right] = \frac{a_0}{2} + \sum_{n=1}^\infty a_n \cos(nx) + \sum_{n=1}^\infty b_n \sin(nx)$$

Calculating $a_0$:

$$a_0 = \frac{1}{\pi} \int_0^\pi (\pi -x) dx = \frac{\pi}{2}$$

Calculating $a_n$:

$$a_n = \frac{1}{\pi} \int_0^\pi (\pi -x) \cos(nx) dx$$

$$=\frac{-1}{\pi} \left( \left[ \frac{x\sin(nx)}{n} \right]_0^\pi - \int_0^\pi \frac{\sin(nx)}{n}dx \right)$$

$$=\frac{1-(-1)^n}{\pi n^2}$$

Calculating $b_n$:

$$b_n = \frac{1}{\pi} \int_0^\pi (\pi -x) \sin(nx) dx$$

$$=\frac{-1}{\pi} \left( \left[ \frac{-(\pi-x)\cos(nx)}{n} \right]_0^\pi - \int_0^\pi \frac{\cos(nx)}{n}dx \right)$$

$$=\frac{1}{n}$$

giving the $FS\left[f(t)\right]$:

$$FS{[f(t)]} = \frac{\pi}{4} + \sum_{n=1}^\infty \frac{1-(-1)^n}{n^2} \cos(nx) + \sum_{n=1}^\infty \frac{\sin(nx)}{n}$$

However I am now stuck on showing:

$$\sum_{n=1}^\infty\frac{1-(-1)^n}{n^2} = \frac{\pi^2}{4}$$

Anyone have any idea? Or see where I might have gone wrong? Kind Regards,

When you compute $a_n$ (which you typo-ed $a_0$), you left off the $1/\pi$. After sticking that back into the FS, and plugging in $0$ you have
$$f(0) =\frac{\pi}{4} +\frac{1}{\pi}\sum \frac{1-(-1)^n}{n^2}.$$
Well, this is a little wrong. The FS doesn't exactly equal the original function. The FS convergence theorem says that the series converges to the average of the left and right hand limits of the function at its discontinuities. So instead of $f(0)$, we should have $(0+\pi)/2$. It's easy to get your result from there.