Solution
= Solution
Set $x=0$ in the convergent <Fourier series> from part (b). Since $f(0)=0$ and every cosine equals one,
$$
0=\frac L2-\frac{4L}{\pi^2}\sum_{k=0}^{\infty}\frac1{(2k+1)^2}.
$$
Rearranging gives
$$
\boxed{\sum_{k=0}^{\infty}\frac1{(2k+1)^2}=\frac{\pi^2}{8}.}
$$
Convergence at the evaluation point is essential when extracting a numerical series identity from a <Fourier series>. Here the function is continuous there and the coefficients are absolutely summable.