Question 7.28: Prove that 1/1² + 1/2² + 1/3² + ··· = π²/6 .
Prove that \frac{1}{1^{2}}+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\cdots=\frac{\pi^{2}}{6}.
We have
\begin{aligned} F(z) & =\frac{\pi \cot \pi z}{z^{2}}=\frac{\pi \cos \pi z}{z^{2} \sin \pi z}=\frac{\left(1-\frac{\pi^{2} z^{2}}{2 !}+\frac{\pi^{4} z^{4}}{4 !}-\cdots\right)}{z^{3}\left(1-\frac{\pi^{2} z^{2}}{3 !}+\frac{\pi^{4} z^{4}}{5 !}-\cdots\right)} \\ & =\frac{1}{z^{3}}\left(1-\frac{\pi^{2} z^{2}}{2 !}+\cdots\right)\left(1+\frac{\pi^{2} z^{2}}{3 !}+\cdots\right)=\frac{1}{z^{3}}\left(1-\frac{\pi^{2} z^{2}}{3}+\cdots\right) \end{aligned}
so that the residue at z=0 is -\pi^{2 / 3}.
Then, as in Problems 7.26 and 7.27,
\oint\limits_{C_{N}} \frac{\pi \cot \pi z}{z^{2}} d z=\sum\limits_{n=-N}^{-1} \frac{1}{n^{2}}+\sum\limits_{n=1}^{N} \frac{1}{n^{2}}-\frac{\pi^{2}}{3}=2 \sum\limits_{n=1}^{N} \frac{1}{n^{2}}-\frac{\pi^{2}}{3}
Taking the limit as N \rightarrow \infty, we have, since the left side approaches zero,
2 \sum\limits_{n=1}^{\infty} \frac{1}{n^{2}}-\frac{\pi^{2}}{3}=0 \quad \text { or } \quad \sum\limits_{n=1}^{\infty} \frac{1}{n^{2}}=\frac{\pi^{2}}{6}
Another Method. Take the limit as a \rightarrow 0 in the result of Problem 7.27. Then, using L’Hospital’s rule,
\underset{a \rightarrow 0}{\lim} \sum\limits_{n=1}^{\infty} \frac{1}{n^{2}+a^{2}}=\sum\limits_{n=1}^{\infty} \frac{1}{n^{2}}=\underset{a \rightarrow 0}{\lim} \frac{\pi a \operatorname{coth} \pi a-1}{2 a^{2}}=\frac{\pi^{2}}{6}