Using partial fraction for $\cot \pi z$ to compute infinite sum

Question

I want to compute the values $\sum_{n=1}^\infty \dfrac{1}{n^2}$ and $\sum_{n=1}^\infty \dfrac{1}{n^4}$ and $\sum_{n=1}^\infty \dfrac{1}{n^6}$ by comparison to the partial fraction development of $\cot \pi z$.

First, I note that $$\pi z\cot(\pi z)=1+2\sum_{n=1}^\infty\dfrac{z^2}{z^2-n^2}=1-2z^2\sum_{n=1}^\infty\dfrac{1}{n^2}\cdot\dfrac{1}{1-\dfrac{z^2}{n^2}}$$

Using the sum of geometric series formula, this equals

$$1-2z^2\sum_{n=1}^\infty\dfrac{1}{n^2}\cdot\left(1+\dfrac{z^2}{n^2}+\dfrac{z^4}{n^4}+\ldots\right)$$

Interchanging the order of summation (since both series are convergent), I get $$1-2z^2\sum_{k=0}^\infty\left(\sum_{n=1}^\infty\dfrac{1}{n^{2(k+1)}}\right)$$

So, I need to find the $z^2,z^4,z^6$ terms of the Taylor expansion of $z\cot(\pi z)$. I can write it as $\dfrac{z\cos(\pi z)}{\sin(\pi z)}$, and both $\cos$ and $\sin$ have Taylor expansions. But how can I get the Taylor expansion for $z\cot(\pi z)$ (in a not-messy way)?

Answer 1

You have lost a few $z^{2k}$ on the way.

$$1-2z^2\sum_{n=1}^\infty\dfrac{1}{n^2}\cdot\left(1+\dfrac{z^2}{n^2}+\dfrac{z^4}{n^4}+\ldots\right) = 1-2z^2\sum_{k=0}^\infty \left(\sum_{n=1}^\infty \frac{1}{n^{2(k+1)}}\right){\color\red{z^{2k}}}.$$

If you only want the very few first terms of the Taylor expansion, dividing the expansions of $\cos$ and $\sin$ is sufficiently non-messy, but going beyond the $z^6$ coefficient starts to be uncomfortable. A technique to make the computation of the Taylor series of $\frac{f(z)}{g(z)}$ less taxing is to make an ansatz

$$\frac{f(z)}{g(z)} = \sum_{n=0}^\infty \frac{c_n}{n!}z^n,$$

and then use the Cauchy product of that with the Taylor series of $g$ to obtain a recursion for the $c_n$.

For the Taylor series of $z\cot z$, we can fiddle a bit to find a representation that yields a nicer recursion,

$$\cot z = \frac{\cos z}{\sin z} = i\frac{e^{iz}+e^{-iz}}{e^{iz}-e^{-iz}} = i + i\frac{2e^{-iz}}{e^{iz}-e^{-iz}} = i + 2i\frac{1}{e^{2iz}-1}.\tag{1}$$

So if we have the Taylor series of $\frac{w}{e^w-1}$, everything follows nicely from that. So we make the ansatz

$$\frac{w}{e^w-1} = \sum_{k=0}^\infty \frac{B_k}{k!}w^k.\tag{2}$$

The $B_k$ are known (famous) as the Bernoulli numbers. From the way we arrived there, it is clear that $\frac{w}{e^w-1}+\frac{w}{2}$ is an even function, hence the only odd term in the series is $-\frac{w}{2}$, and $B_{2k+1} = 0$ for $k > 0$.

Multiplying $(2)$ with the Taylor series of $e^w-1$ produces

$$\begin{align} w &= \left(\sum_{k=0}^\infty \frac{B_k}{k!}w^k\right)\left(\sum_{m=1}^\infty \frac{w^m}{m!}\right)\\ &= \sum_{n=1}^\infty \left(\sum_{k=0}^{n-1}\frac{B_k}{k!(n-k)!}\right)w^n\\ &= \sum_{n=1}^\infty \left(\sum_{k=0}^{n-1}\binom{n}{k}B_k\right)\frac{w^n}{n!}. \end{align}$$

Thus we have $B_0 = 1$ and the relation

$$\sum_{k=0}^{n-1}\binom{n}{k}B_k = 0\tag{3}$$

for $n > 1$. It is still not trivial (in terms of amount of computation) to compute the $B_{2k}$ for larger $k$, but it is much easier and less work than dividing the Taylor series of $\cos$ and $\sin$.

I'll leave it to you to connect $(1)$ and $(2)$ to obtain the Taylor series of $\pi z\cot \pi z$ in terms of the Bernoulli numbers, and to compute the first Bernoulli numbers with the relation $(3)$.