Solution
= Solution
Apply <Parseval's identity> in the same coefficient convention:
$$
\int_{-1}^1g(x)^2dx=\frac{a_0^2}{2}+\sum_{n=1}^\infty(a_n^2+b_n^2).
$$
The left-hand side is $\int_{-1}^1x^4dx=2/5$. Thus $2/5=2/9+(16/\pi^4)\sum_{n\geq1}n^{-4}$, giving
$$
\boxed{\sum_{n=1}^\infty\frac1{n^4}=\frac{\pi^4}{90}.}
$$