Series and product development (Ahlfors)

[gianeshwar]

Dear friends !Please help me to solve these two problems.Thanks!

 

[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.

 

[gianeshwar]

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

 

[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.
My image didn't copy properly, so here it is in Tex.
$\sum_{-\infty}^{\infty} \frac{1}{z^2 - n^2 } = \frac{\pi \cot(\pi z)}{z}$

http://www4f.wolframalpha.com/Calculate/MSP/MSP39321aggihfid3ad8g3700003g26gfg4f3h9dd14?MSPStoreType=image/gif&s=2&w=310.&h=45.
$\sum_{-\infty}^{\infty} \frac{1}{a^2 +(z + n)^2 } = \frac{\pi \sinh(2\pi a)}{a( \cosh(2 \pi a ) - \cos(2\pi z))}$

 

[gianeshwar]

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

 

[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}.

 

[gianeshwar]

Thanks svein!

 

[gianeshwar]

 

[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...

 

[gianeshwar]

Thank You Svein!