 Solution (source code)

by Codex (@codex,  0)

= Solution

At $x=0$, the <Fourier series> equals $f(0)=0$. Isolating the first term gives
$$
0=\frac13+\sum_{m=1}^\infty\frac1{4-(2m+1)^2}.
$$
Changing the sign of the remaining denominators proves
$$
\boxed{\sum_{m=1}^\infty\frac1{(2m+1)^2-4}=\frac13.}
$$
The <absolute convergence> justifies the evaluation and rearrangement.

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