Week 11 Worksheet - Solutions

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Summing Series

1. We take $f:\mathbb{C}\setminus \{0\}\to \mathbb{C}$ to be $f(z)={\displaystyle \frac{1}{z^4}}$. Recall that $g(z)={\displaystyle \frac{\sin (\pi z)}{\cos (\pi z)}}$ is uniformly bounded on ${\mathrm{\Gamma }}_N$. We can estimate that $$\left|\underset{{\mathrm{\Gamma }}_N}{\int }fg\right|\le \ell ({\mathrm{\Gamma }}_N){\displaystyle \frac{M}{N^4}}\le {\displaystyle \frac{16M}{N^3}}$$ because $|z|\ge N$ for all $z\in {\mathrm{\Gamma }}_N$. Thus the contour integral converges to zero as $N\to \mathrm{\infty }$. Cauchy's residue theorem gives $$\underset{{\mathrm{\Gamma }}_N}{\int }fg=2\pi i\sum _{n=-N}^N\mathsf{Res}(fg,n)\mathsf{wind}({\mathrm{\Gamma }}_N,n)$$ because the only poles of $fg$ are at $n\in \{-N,\dots ,N\}$. As usual $$\mathsf{Res}(fg,n)={\displaystyle \frac{1}{\pi n^4}}$$ for all non-zero $n$ and it reamins to calculate the residue at zero. The order of the pole at zero is five. From our lemma $$\mathsf{Res}(fg,0)=\underset{z\to 0}{lim}{\displaystyle \frac{1}{4!}}{\left({\displaystyle \frac{z^5\cos (\pi z)}{z^4\sin (\pi z)}}\right)}^{(4)}=\underset{z\to 0}{lim}{\displaystyle \frac{1}{4!}}{({\displaystyle \frac{z}{\sin (\pi z)}}\cos (\pi z))}^{(4)}$$ and the power series $${\displaystyle \frac{w}{\sin w}}=1+{\displaystyle \frac{w^2}{6}}+{\displaystyle \frac{w^4}{240}}+\cdots$$ we get $$\begin{array}{rl}\mathsf{Res}(fg,0)& =\underset{z\to 0}{lim}{\displaystyle \frac{1}{4!}}{\left(({\displaystyle \frac{1}{\pi }}+{\displaystyle \frac{\pi }{6}}{z}^2+{\displaystyle \frac{7{\pi }^3}{360}}{z}^4-\cdots )(1-{\displaystyle \frac{{\pi }^2}{2}}{z}^2+{\displaystyle \frac{{\pi }^4}{24}}{z}^4-\cdots )\right)}^{(4)}\\ & =\underset{z\to 0}{lim}{\displaystyle \frac{1}{4!}}{({\displaystyle \frac{1}{\pi }}-{\displaystyle \frac{\pi }{3}}{z}^2-{\displaystyle \frac{{\pi }^3}{45}}{z}^4+\cdots )}^{(4)}\\ & =-{\displaystyle \frac{{\pi }^3}{45}}\end{array}$$ and we can put everything together. Cauchy's residue theorem gives $$2\pi i(2\sum _{n=1}^N{\displaystyle \frac{1}{\pi n^4}}-{\displaystyle \frac{{\pi }^3}{45}})=\underset{{\mathrm{\Gamma }}_N}{\int }fg$$ and the limit as $N\to \mathrm{\infty }$ gives ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}{\displaystyle \frac{1}{n^4}}={\displaystyle \frac{{\pi }^4}{90}}}$
2. We would want to use $f(z)=1/z^3$ but with that choice $$\mathsf{Res}(fg,-n)=-{\displaystyle \frac{1}{\pi n^3}}=-\mathsf{Res}(fg,n)$$ for all $n\in \mathbb{N}$ and Cauchy's residue theorem only gives $$\underset{{\mathrm{\Gamma }}_N}{\int }fg=2\pi i\mathsf{Res}(fg,0)$$ which does not involve ${\displaystyle \sum _{n=1}^N{\displaystyle \frac{1}{n^3}}}$