- Setting \(\tau = 1 / 2\) in the expression
\[ \sum_ {m = - \infty} ^ {\infty} \frac {1}{(m + \tau) ^ {2}} = \frac {\pi^ {2}}{\sin^ {2} (\pi \tau)}, \]
deduce that
\[ \sum_ {m \geq 1, m \text {odd}} \frac {1}{m ^ {2}} = \frac {\pi^ {2}}{8} \quad \text {and} \quad \sum_ {m \geq 1} \frac {1}{m ^ {2}} = \frac {\pi^ {2}}{6} = \zeta (2). \]
Similarly, using \(\textstyle \sum { 1 / ( m + \tau ) ^ { 4 } }\) deduce that
\[ \sum_ {m \geq 1, m \text {odd}} \frac {1}{m ^ {4}} = \frac {\pi^ {4}}{9 6} \quad \text {and} \quad \sum_ {m \geq 1} \frac {1}{m ^ {4}} = \frac {\pi^ {4}}{9 0} = \zeta (4). \]
These results were already obtained using Fourier series in the exercises at the end of Chapters 2 and 3 in Book I.
Solution
At \(\tau=1/2\), \[ \sum_{m\in\mathbb Z}\frac1{(m+1/2)^2}=\pi^2. \] Since \((m+1/2)^{-2}=4(2m+1)^{-2}\) and the positive and negative odd integers contribute equally, \[ \pi^2=8\sum_{\substack{r\ge1\\r\text{ odd}}}\frac1{r^2}, \] so \[ \sum_{r\ge1,\ r\text{ odd}}r^{-2}=\frac{\pi^2}{8}. \] Now \[ \zeta(2)=\sum_{r\text{ odd}}r^{-2}+\sum_{r\ge1}(2r)^{-2} =\frac{\pi^2}{8}+\frac14\zeta(2), \] whence \(\zeta(2)=\pi^2/6\).
Differentiate the identity \[ \sum_{m\in\mathbb Z}(m+\tau)^{-2}=\pi^2\csc^2(\pi\tau) \] twice. Since \[ \frac{d^2}{d\tau^2}(m+\tau)^{-2}=6(m+\tau)^{-4}, \] and, writing \(u=\pi\tau\), \[ \frac{d^2}{d\tau^2}\bigl(\pi^2\csc^2u\bigr) =\pi^4\bigl(4\csc^2u\cot^2u+2\csc^4u\bigr), \] at \(\tau=1/2\) we get \[ 6\sum_{m\in\mathbb Z}(m+1/2)^{-4}=2\pi^4. \] Therefore \[ \sum_{m\in\mathbb Z}(m+1/2)^{-4}=\frac{\pi^4}{3}. \] Since this equals \(32\sum_{r\ge1,\ r\text{ odd}}r^{-4}\), \[ \sum_{r\ge1,\ r\text{ odd}}r^{-4}=\frac{\pi^4}{96}. \] Finally \[ \zeta(4)=\frac{\pi^4}{96}+\frac1{16}\zeta(4), \] so \(\zeta(4)=\pi^4/90\).