= Solution
Here $f(x)^2=x^2$, so
$$
\frac1L\int_{-L}^Lf(x)^2dx=\frac{2L^2}{3}.
$$
Substituting the coefficients from part (b) into <Parseval's identity> gives
$$
\frac{2L^2}{3}=\frac{L^2}{2}
+\frac{16L^2}{\pi^4}\sum_{k=0}^{\infty}\frac1{(2k+1)^4}.
$$
Hence the sum over \i[all] positive odd integers is $\pi^4/96$. The requested series starts at $n=1$, whose first denominator is $3^4$, so it excludes the contribution $1^{-4}$:
$$
\boxed{\sum_{n=1}^{\infty}\frac1{(2n+1)^4}=\frac{\pi^4}{96}-1.}
$$
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