= Solution
Write $h_n=a_n\cos(n\pi x/L)+b_n\sin(n\pi x/L)$. For the partial sum $S_N=a_0/2+\sum_{n=1}^Nh_n$, ordinary multiplication gives
$$
S_N^2=\frac{a_0^2}4+a_0\sum_{n=1}^Nh_n+\sum_{m,n=1}^Nh_mh_n.
$$
On $[-L,L]$, the integrals of every nonconstant sine or cosine vanish. <Fourier orthogonality> further gives $\int\cos(m\pi x/L)\cos(n\pi x/L)dx=L\delta_{mn}$ and the same identity for two sines, while every sine-cosine integral is zero. Thus
$$
\frac1{2L}\int_{-L}^LS_N^2dx=\frac{a_0^2}4+\frac12\sum_{n=1}^N(a_n^2+b_n^2).
$$
For square-integrable $f$, its <Fourier partial sums> converge to $f$ in <L2 space>, so their squared norms converge to the squared norm of $f$. Passing to the limit proves <Parseval's identity>:
$$
\boxed{\frac1T\int_{-T/2}^{T/2}f(x)^2dx=\left(\frac{a_0}2\right)^2+\frac12\sum_{n=1}^{\infty}(a_n^2+b_n^2).}
$$
Working with finite partial sums before taking the norm limit avoids an unjustified interchange of an infinite double series and integration.
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