= Solution
{c}
Set $p=3$ and $x=0$ in the <Fourier series> from <b>. Since the <function> is continuous there, <pointwise convergence of a piecewise smooth Fourier series> gives
$$
0=\frac13-\frac16-6\!\sum_{\substack{n\geq1\\n\ne3}}
\frac{(-1)^n}{n^2-9}.
$$
It follows that
$$
\sum_{\substack{n\geq1\\n\ne3}}\frac{(-1)^n}{n^2-9}=\frac1{36}.
$$
The $n=1$ and $n=2$ terms are $1/8$ and $-1/5$, respectively. Subtracting them gives the required absolutely convergent <series>:
$$
\boxed{\sum_{n=4}^\infty\frac{(-1)^n}{n^2-9}
=\frac1{36}-\frac18+\frac15=\frac{37}{360}}.
$$
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