= Solution
{c}
By <Dirichlet's theorem on Fourier series>, a piecewise continuously differentiable <periodic function> has a <Fourier series> converging at each point to the average of its one-sided limits. In particular it converges to the <function> at every point of <continuity>. Our <periodic extension> satisfies these hypotheses, and $x=\pi/2$ is a point of <continuity> with $f(\pi/2)=\pi^2/4$.
At this point $\sin(n\pi/2)=0$ for even $n$. For $n=2j-1$,
$$
\sin\frac{(2j-1)\pi}{2}=(-1)^{j+1},\qquad
b_{2j-1}=\frac{2\pi}{2j-1}-\frac8{\pi(2j-1)^3}.
$$
Substituting in the <Fourier series> from part (c) proves
$$
\boxed{\frac{\pi^2}{4}=\sum_{j=1}^{\infty}(-1)^{j+1}\left[\frac{2\pi}{2j-1}-\frac8{\pi(2j-1)^3}\right].}
$$
The sum is taken in its displayed order. Discarding its zero even-indexed terms is valid and does not rearrange the remaining terms.
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