$\frac{1}{z} \prod_{n=1}^{\infty} \frac{n^2}{n^2 - z^2} = \frac{1}{z} + 2z\sum_{n=1}^{\infty} \frac{(-1)^n}{z^2-n^2}$?

Question

I am trying to show that

$$\frac{1}{z} \prod_{n=1}^{\infty} \frac{n^2}{n^2 - z^2} = \frac{1}{z} + 2z\sum_{n=1}^{\infty} \frac{(-1)^n}{z^2-n^2}$$

This question stems from the underlying homework problem, which asks to show $$ \frac{\pi}{\sin(\pi z)} = \frac{1}{z} + 2z\sum_{n=1}^{\infty} \frac{(-1)^n}{z^2-n^2}, $$ to which I am at my wits end. I have a couple of identities on hand, namely $$ \pi \cot (\pi z) = \frac{1}{z} + \sum_{n \in \mathbb{Z}; n \neq 0} \frac{1}{z - n} + \frac{1}{n} $$ and $$ \frac{\sin (\pi z)}{\pi} = z \prod_{n=1}^{\infty} \left( 1 - \frac{z^2}{n^2} \right) $$ and $$ \frac{\pi^2}{\sin^2 (\pi z)} = \sum_{n \in \mathbb{Z}} \frac{1}{(z - n)^2} $$ I've tried fooling around with these identities and am getting nowhere. Any hints or suggestions would be greatly appreciated.

score 3 · Answer 1 · edited Apr 13 '17 at 12:21

In this answer, it is derived that $$ \begin{align} \pi\cot(\pi z) &=\sum_{k=-\infty}^\infty\frac{1}{z+k}\\ &=\frac1z+\sum_{k=1}^\infty\frac{2z}{z^2-k^2}\tag{1} \end{align} $$ To get the alternating series, note that $$ \begin{align} \frac1z+2z\sum_{k=1}^\infty\frac{(-1)^k}{z^2-k^2} &=\sum_{k=-\infty}^\infty\frac{(-1)^k}{z+k}\\ &=\sum_{k=-\infty}^\infty\frac{2}{z+2k}-\frac1{x+k}\\ &=\sum_{k=-\infty}^\infty\frac{1}{z/2+k}-\frac1{x+k}\\[6pt] &=\pi\cot(\pi z/2)-\pi\cot(\pi z)\\[7pt] &=\pi\frac{1+\cos(\pi z)}{\sin(\pi z)}-\pi\frac{\cos(\pi z)}{\sin(\pi z)}\\ &=\frac\pi{\sin(z)}\tag{2} \end{align} $$

Ethan Splaver · Accepted Answer · 2013-10-22T02:33:07.787

2

$$\sum_{n=1}^\infty\frac{(-1)^n}{z^2-n^2}=2\sum_{n=1}^\infty\frac{1}{z^2-(2n)^2}-\sum_{n=1}^\infty\frac{1}{z^2-n^2}$$

$$=\frac{1}{2}\sum_{n=1}^\infty\frac{1}{(z/2)^2-n^2}-\sum_{n=1}^\infty\frac{1}{z^2-n^2}$$

Now since,

$$\pi \cot(\pi z)=\frac{1}{z}+2z\sum_{n=1}^\infty\frac{1}{z^2-n^2}$$

We get that, $$\pi\coth(\frac{\pi z}{2})=\frac{2}{z}+2z\frac{1}{2}\sum_{n=1}^\infty\frac{1}{(z/2)^2-n^2}$$

And so, $$\pi\cot(\frac{\pi z}{2})-\pi \cot(\pi z)=\frac{1}{z}+2z(\frac{1}{2}\sum_{n=1}^\infty\frac{1}{(z/2)^2-n^2}-\sum_{n=1}^\infty\frac{1}{z^2-n^2}) $$ $$=\frac{1}{z}+2z\sum_{n=1}^\infty \frac{(-1)^n}{z^2-n^2}$$

Now since $$\cot(\frac{\pi z}{2})-\cot(\pi z)=\frac{1}{\sin(\pi z)}$$

We get that:

$$\frac{\pi}{\sin(\pi z)}=\frac{1}{z}+2z\sum_{n=1}^\infty \frac{(-1)^n}{z^2-n^2}$$

As required

edited Oct 22 '13 at 02:33

answered Oct 22 '13 at 00:42

Ethan Splaver

10,812

Where the expansion for $\pi \cot(\pi z)$ can be obtained by taking the logarithmic derivative of the product formula for the sine function, or by re-indexing the sum over the integers you gave as the partial fraction decomposition for the cotangent function. – Ethan Splaver Oct 22 '13 at 00:45
I really wish I could upvote this more times. Thank you so much! – tylerc0816 Oct 22 '13 at 01:06
1

Do you really mean $\coth$ or is that a typo? – robjohn Oct 22 '13 at 01:19
@robjohn Fixed thanks – Ethan Splaver Oct 22 '13 at 02:34

score 1 · Answer 3 · answered Oct 22 '13 at 00:46

Let $ \displaystyle f(z) = \frac{\pi}{\sin \pi z} - \frac{1}{z}$.

Then according to the Mittag-Leffler pole expansion theorem, $$ \frac{\pi}{\sin \pi z} - \frac{1}{z} = f(0) + \sum_{n=1}^{\infty} \text{Res}[f,n] \Big( \frac{1}{z-n} + \frac{1}{n} \Big) + \sum_{n=1}^{\infty} \text{Res}[f,-n] \Big( \frac{1}{z+n} - \frac{1}{n} \Big)$$

$$ = \sum_{n=1}^{\infty} (-1)^{n} \Big( \frac{1}{z-n} + \frac{1}{n} \Big) + \sum_{n=1}^{\infty} (-1)^{n} \Big( \frac{1}{z+n} - \frac{1}{n} \Big)$$

$$ = \sum_{n=1}^{\infty} (-1)^{n} \Big(\frac{1}{z-n} + \frac{1}{z+n} \Big) = \sum_{n=1}^{\infty} (-1)^{n} \frac{2z}{z^{2}-n^{2}}$$

$\frac{1}{z} \prod_{n=1}^{\infty} \frac{n^2}{n^2 - z^2} = \frac{1}{z} + 2z\sum_{n=1}^{\infty} \frac{(-1)^n}{z^2-n^2}$?

3 Answers3