Assume that $$S(x) = \sum_{n=1}^{\infty}B_n\sin nx$$ is the sum of the Fourier sine series on $[0, \pi]$ for the function $f(x)$ that is equal to $1$ on the interval $[0, \frac{\pi}{2})$ and $0$ on $[\frac{\pi}{2},\pi]$
Evalauate the sum $$\sum_{n=1}^{\infty}{B_n}^2$$ Hint: The fourier sine series is equal to the full Fourier series of the odd extension of $f$
So with a bit of elbow grease, one would get $$B_n = \frac{2}{n\pi}$$ I believe the idea of the problem is to not use the famous sum for $\frac{1}{n^2}$. However, this hint has me lost, since our extension was odd, don't the $A_n$ cosine Fourier coefficients disappear? I computed them and they did so. I am not sure what information this hint gives me. I was thinking about perhaps trying to rewrite the problem in a way that yields Parseval's identity, but I cannot think of anything clever (or obvious) which would get the proper sum indexing.