Let $f(z)=1 /\left(e^{z}-1\right)$. Find the first three terms in the Laurent expansion for $f(z)$ valid for $0<|z|<2 \pi$.

Now let $n$ be a positive integer, and define

$$\begin{aligned} &f_{1}(z)=\frac{1}{z}+\sum_{r=1}^{n} \frac{2 z}{z^{2}+4 \pi^{2} r^{2}} \\ &f_{2}(z)=f(z)-f_{1}(z) \end{aligned}$$

Show that the singularities of $f_{2}$ in $\{z:|z|<2(n+1) \pi\}$ are all removable. By expanding $f_{1}$ as a Laurent series valid for $|z|>2 n \pi$, and $f_{2}$ as a Taylor series valid for $|z|<2(n+1) \pi$, find the coefficients of $z^{j}$ for $-1 \leq j \leq 1$ in the Laurent series for $f$ valid for $2 n \pi<|z|<2(n+1) \pi$.

By estimating an appropriate integral around the contour $|z|=(2 n+1) \pi$, show that

$$\sum_{r=1}^{\infty} \frac{1}{r^{2}}=\frac{\pi^{2}}{6}$$