# Series and product development (Ahlfors)

1. Jun 15, 2015

### gianeshwar

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2. Jun 15, 2015

### RUber

Hello. It seems like these series are related to p-series, with n^2 it is more like the Basel problem (https://en.wikipedia.org/wiki/Basel_problem). I do not immediately see the proper conversion, but I will continue to think about it.

3. Jun 15, 2015

### gianeshwar

Thank you very much RUber.I was struggling with these problems after studying Jensons formulae and Mittag Lefflers theorems.

4. Jun 15, 2015

### RUber

I had to ask wolframalpha.com, but was provided with the following solutions. These types of problems usually require a great deal of manipulation to show, even when the answer is known.

http://www4f.wolframalpha.com/Calculate/MSP/MSP652520diec13gdedd2dh00003i3bedh373981c66?MSPStoreType=image/gif&s=20&w=168.&h=45. [Broken]
My image didn't copy properly, so here it is in Tex.
$\sum_{-\infty}^{\infty} \frac{1}{z^2 - n^2 } = \frac{\pi \cot(\pi z)}{z}$

$\sum_{-\infty}^{\infty} \frac{1}{a^2 +(z + n)^2 } = \frac{\pi \sinh(2\pi a)}{a( \cosh(2 \pi a ) - \cos(2\pi z))}$

Last edited by a moderator: May 7, 2017
5. Jun 15, 2015

### gianeshwar

Thanks RUber! I have progressed a little further.Will share soon.

6. Jun 16, 2015

### Svein

A bit further up on the page, marked as (11) it says: $\pi\cot(\pi z)=\frac{1}{z}+\sum_{n=1}^{\infty}\frac{2z}{z^{2}-n^{2}}$. Keep that in mind. Now $\sum_{-\infty}^{\infty}\frac{1}{z^{2}-n^{2}}=\sum_{n=-\infty}^{-1}\frac{1}{z^{2}-n^{2}}+\frac{1}{z^{2}}+\sum_{n=1}^{\infty}\frac{1}{z^{2}-n^{2}}$. Since (-n)2=n2, we have $\sum_{-\infty}^{\infty}\frac{1}{z^{2}-n^{2}}=\frac{1}{z^{2}}+2 \cdot\sum_{n=1}^{\infty}\frac{1}{z^{2}-n^{2}}$. Now we are almost there - divide the expression in (11) by z (remembering to keep away from 0), you have $\frac{\pi\cot(\pi z)}{z}=\frac{1}{z^{2}}+2\cdot\sum_{n=1}^{\infty}\frac{1}{z^{2}-n^{2}}$. Conclusion: $\sum_{-\infty}^{\infty}\frac{1}{z^{2}-n^{2}}=\frac{\pi\cot(\pi z)}{z}$.

7. Jun 16, 2015

### gianeshwar

Thanks svein!

8. Jun 16, 2015

### gianeshwar

9. Jun 17, 2015

### Svein

I shall give you some hints:
1. $\frac{1}{(z+n)^{2}+a^{2}}=\frac{1}{2ia}(\frac{1}{z+n-ia}-\frac{1}{z+n-ia})$
2. $\sum_{n=-\infty}^{\infty}\frac{1}{z+n+ia}=\sum_{n=-\infty}^{\infty}\frac{1}{z-n+ia}$
3. Let w = z+ia. We already know that $\sum_{n=-\infty}^{\infty}\frac{1}{w-n}=\pi\cot\pi w$.
4. Do the same thing with w = z-ia.
5. Substituting back, you get $\sum_{-\infty}^{\infty}\frac{1}{(z+n)^{2}+a^{2}}=\frac{1}{2ia}(\pi\cot\pi (z+ia)-\pi\cot\pi (z-ia))$
The rest is left to the student...

10. Jun 17, 2015

### gianeshwar

Thank You Svein!