Sum of series $\sum_{k=1}^{\infty} \frac{1}{k^2}=\frac{\pi^2}{6}$

Question

I first derived the Fourier series for $f(x) = abs(x)$ to be: $\frac{\pi}{2} - \frac{4}{\pi} \sum_{k=1}^{\infty} \frac{cos((2k-1)x)}{(2k-1)^2}$

Now, I want to use this at x=0 to find $\sum_{k=1}^{\infty} \frac{1}{k^2}$. This is my work:

$abs(x) = 0 = \frac{\pi}{2}- \frac{4}{\pi} \sum_{k=1}^{\infty} \frac{1}{(2k-1)^2}$

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

$\sum_{k=1}^{\infty} \frac{1}{(2k-1)^2} = \frac{\pi^2}{8}$

I know the answer should be $\frac{\pi^2}{6}$, not $\frac{\pi^2}{8}$, but I am not sure where I am going wrong. Is my Fourier series incorrect?

Edit: How can I go from $\sum_{k=1}^{\infty} \frac{1}{(2k-1)^2} = \frac{\pi^2}{8}$ to $\sum_{k=1}^{\infty} \frac{1}{k^2}=\frac{\pi^2}{6}$?

Answer 1

Let $A=\sum_{k=1}^{\infty}{\frac{1}{k^2}}$. Then we can see that $$\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\cdots=\sum_{k=1}^{\infty}{\frac{1}{(2k)^2}}=\frac{1}{4}\sum_{k=1}^{\infty}{\frac{1}{k^2}}=\frac{A}{4}.$$ It then follows that $$\frac{\pi^2}{8}=\sum_{k=1}^{\infty}{\frac{1}{(2k-1)^2}}=\sum_{k=1}^{\infty}{\frac{1}{k^2}}-\sum_{k=1}^{\infty}{\frac{1}{(2k)^2}}=A-\frac{A}{4}=\frac{3}{4}A,$$ so $A=\pi^2/6$.