How to find $1+1/2^2+1/3^2+...$ using Fourier Series.

Question

How to find $1+1/2^2+1/3^2+...$ using Fourier Series. I have already worked out that the Fourier cosine series for $f(x) = x$ is:

$x = \frac{4L}{\pi^2} (1+\frac{1}{3^2}+\frac{1}{5^2} \cdots) -\frac{4L}{\pi^2}(\cos{\frac{\pi x}{L}}+...+\frac{1}{n^2}\cos{n \pi x/L}+...)$ where n is odd.

hint given is to evaluate this expression at $x = 0$. But then I just get $0 = 0$?

Edit, I have already worked out $1+1/3^2+1/5^2+... = \pi^2/8$

Answer 1

If you proved $$ \sum_{k\geq 0}\frac{1}{(2k+1)^2}=\frac{\pi^2}{8},$$ the remaining part is easy. Just notice that: $$ \zeta(2)=\sum_{n\geq 1}\frac{1}{n^2}=\sum_{k\geq 0}\frac{1}{(2k+1)^2}+\sum_{k\geq 1}\frac{1}{(2k)^2} = \frac{\pi^2}{8}+\frac{\zeta(2)}{4},$$ hence $\frac{3}{4}\zeta(2)=\frac{\pi^2}{8}$ gives: $$ \zeta(2)=\frac{\pi^2}{6}$$ as wanted.

Answer 2

Let $f(x)=x$ over the interval $x \in (-\pi,\pi)$. The Fourier series of this function is

Using Parseval's identity we obtain,

Where

Thus, $|a_n|^2=\frac{1}{n^2}$.

and,

Q.E.D