Find the sum of the series \(\displaystyle\sum_{n=1}^{\infty}\frac{\sin((2n-1)x)}{2n-1}\) on \((0,2\pi)\).
Show that \(\displaystyle\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}=\frac{\pi^2}{8}\).
Solution
1. Identify the odd sine series as the Fourier series of a square wave.
Define the \(2\pi\)-periodic function \[ q(x)= \begin{cases} 1,&0<x<\pi,\\ -1,&\pi<x<2\pi, \end{cases} \] with arbitrary values at integer multiples of \(\pi\). Its cosine coefficients vanish, and its sine coefficients are \[ b_k=\frac1\pi\int_0^{2\pi}q(x)\sin(kx)\,dx. \] For even \(k\), \(b_k=0\). For odd \(k\), \[ b_k=\frac4{\pi k}. \] Hence the Fourier series of \(q\) is \[ q(x)\sim \frac4\pi\sum_{n=1}^\infty \frac{\sin((2n-1)x)}{2n-1}. \]
The function \(q\) is piecewise \(C^1\), so the Dirichlet convergence theorem gives convergence at every point to the midpoint of the one-sided limits. Therefore, for \(0<x<2\pi\), \[ \sum_{n=1}^\infty\frac{\sin((2n-1)x)}{2n-1} = \begin{cases} \dfrac\pi4,&0<x<\pi,\\[4pt] 0,&x=\pi,\\[4pt] -\dfrac\pi4,&\pi<x<2\pi. \end{cases} \]
2. Apply Parseval’s identity.
Since \(q^2=1\) almost everywhere, \[ \int_0^{2\pi}|q(x)|^2\,dx=2\pi. \] Parseval’s identity for the real Fourier coefficients gives \[ \frac1\pi\int_0^{2\pi}|q(x)|^2\,dx =\sum_{k=1}^\infty b_k^2, \] because the constant and cosine coefficients vanish. Thus \[ 2 =\sum_{n=1}^\infty \left(\frac4{\pi(2n-1)}\right)^2 =\frac{16}{\pi^2} \sum_{n=1}^\infty\frac1{(2n-1)^2}. \] Therefore \[ \boxed{ \sum_{n=1}^\infty\frac1{(2n-1)^2}=\frac{\pi^2}{8}.} \]