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## Series Contest (SC3)

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Moderators: Random Variable, sos440

### Re: Series Contest (SC3)

Tue Oct 15, 2013 3:24 pm

Posts: 852
Location: Jaipur, India

Random Variable wrote:
Are you sure that's the answer to problem 6?

According to that formula we need the singular value $\displaystyle k_{25}$ to evaluate $\displaystyle \sum_{n=-\infty}^\infty e^{-5\pi n^2}$.

According to Wolfram Mathworld, $\displaystyle K(k_{25}) = \frac{\sqrt{5}+2}{20} \frac{\Gamma^{2} (\frac{1}{4})}{\sqrt{\pi}}$

http://mathworld.wolfram.com/EllipticIn ... Value.html

So $\displaystyle K(k_{25}) = \frac{\sqrt{5}+2}{20} \frac{\Gamma^{2} (\frac{1}{4})}{\sqrt{\pi}} = \frac{\pi}{2} \theta^{2} (e^{- 5 \pi} ) = \frac{\pi}{2} \Big( \sum_{k=-\infty}^{\infty} e^{- 5 \pi n^{2}} \Big)^{2}$

$\displaystyle \implies \sum_{k=-\infty}^{\infty} e^{- 5 \pi n^{2}} = \sqrt{\frac{\sqrt{5}+2}{10}} \frac{\Gamma (\frac{1}{4})}{\pi^{3/4}} = \sqrt{\frac{\sqrt{5}+2}{10}} \frac{\pi \csc (\frac{\pi}{4})}{\Gamma( \frac{3}{4}) \pi^{3/4}}$

$\displaystyle = \sqrt{\frac{\sqrt{5}+2}{5}} \frac{\pi^{\frac{1}{4}}}{\Gamma (\frac{3}{4})}$

Thanks for pointing out the typo. I will fix it right now.

### Re: Series Contest (SC3)

Wed Oct 16, 2013 2:38 am

Posts: 852
Location: Jaipur, India

Problem 7

Prove that

$\displaystyle \sum_{n=-\infty}^{\infty} \frac{\Gamma(a+n)\Gamma(b+n)}{\Gamma(c+n)\Gamma(d+n)}=\frac{\pi^2}{\sin(\pi a)\sin(\pi b)}\frac{\Gamma(c+d-a-b-1)}{\Gamma(c-a)\Gamma(c-b)\Gamma(d-a)\Gamma(d-b)}$

where $\text{Re}(c+d-a-b)>1$.

This is known as 'Dougall's Bilateral Sum'.

### Re: Series Contest (SC3)

Fri Oct 25, 2013 7:31 pm
sos440
Integration Guru

Posts: 124
Location: California, US

Shobhit wrote:
Problem 7

Prove that

$\displaystyle \sum_{n=-\infty}^{\infty} \frac{\Gamma(a+n)\Gamma(b+n)}{\Gamma(c+n)\Gamma(d+n)}=\frac{\pi^2}{\sin(\pi a)\sin(\pi b)}\frac{\Gamma(c+d-a-b-1)}{\Gamma(c-a)\Gamma(c-b)\Gamma(d-a)\Gamma(d-b)}$

This is known as 'Dougall's Bilateral Sum'.

It seems somehow related to the Ramanujan's beta integral. Interesting...
Every integral harbors a story as fascinating as a comic book!

### Re: Series Contest (SC3)

Sat Oct 26, 2013 5:30 pm

Posts: 852
Location: Jaipur, India

sos440 wrote:
It seems somehow related to the Ramanujan's beta integral. Interesting...

Yes, that is what came to my mind when I saw it.

The idea is to integrate $\displaystyle f(z) = \frac{\Gamma(a+z)\Gamma(b+z)}{\Gamma(c+z)\Gamma(d+z)}\pi \cot(\pi z)$ around a large circle centered at the origin and then use the Residue Theorem.

Have a look at page no 128 of

Note - Here, $\Pi(z)$ has been used to denote gamma function instead of $\Gamma(z)$.

### Re: Series Contest (SC3)

Mon Nov 04, 2013 11:12 pm

Posts: 138
Location: North Londinium, UK
Shobhit wrote:
Problem 7

Prove that

$\displaystyle \sum_{n=-\infty}^{\infty} \frac{\Gamma(a+n)\Gamma(b+n)}{\Gamma(c+n)\Gamma(d+n)}=\frac{\pi^2}{\sin(\pi a)\sin(\pi b)}\frac{\Gamma(c+d-a-b-1)}{\Gamma(c-a)\Gamma(c-b)\Gamma(d-a)\Gamma(d-b)}$

This is known as 'Dougall's Bilateral Sum'.

I've a proof of this in one of my books, but haven't tried it myself... Yet. It's worth noting, however, that the case $\displaystyle d=1$ is equivalent to the hypergeometric function, since $\displaystyle 1/\Gamma(n)=0$ for negative integers. This explains why the term $\displaystyle 1/n!$ is introduced into hypergeometric series.

### Re: Series Contest (SC3)

Fri Nov 08, 2013 1:04 pm

Posts: 852
Location: Jaipur, India

Problem 8

Prove that

$\displaystyle \frac{4K^2}{\pi^2}=\sum_{j=0}^\infty \frac{\left(\frac{1}{2} \right)_j^3}{(j!)^3}(2kk')^{2j}={\;}_3F_2 \left( \frac{1}{2},\frac{1}{2},\frac{1}{2};1,1;(2kk')^2\right)$

All symbols have their standard meaning in the theory of Elliptic Functions.

[BONUS]

$\displaystyle \frac{4}{\pi}=\sum_{n=0}^\infty \frac{\left(\frac{1}{2} \right)_n^3}{(n!)^3}\frac{6n+1}{4^n}$

[NOTE] $\displaystyle (a)_n = \frac{\Gamma(a+n)}{\Gamma(a)}$

### Re: Series Contest (SC3)

Wed Dec 11, 2013 9:29 am

Posts: 38
Location: India, West Bengal
Shobhit wrote:
Problem 8$\displaystyle \frac{4K^2}{\pi^2}=\sum_{j=0}^\infty \frac{\left(\frac{1}{2} \right)_j^3}{(j!)^3}(2kk')^{2j}={\;}_3F_2 \left( \frac{1}{2},\frac{1}{2},\frac{1}{2};1,1;(2kk')^2\right)$

I had my eye on that problem for some time.
$\displaystyle K(k) = \frac{\pi}{2} {\;}_2F_1 \left (\frac{1}{2}, \frac{1}{2}; 1; k^2 \right )$

Using Kummer's transformation, we have

$\displaystyle K(k) = \frac{\pi}{2} {\;}_2F_1 \left ( \frac{1}{4}, \frac{1}{4}; 1; 4k^2k'^2 \right )$

By the Clausen Transform,

$\displaystyle \left [ {\;}_2F_1 \left ( \frac{1}{4}, \frac{1}{4}; 1; 4k^2k'^2 \right ) \right ]^2 = {\;}_3F_2 \left( \frac{1}{2},\frac{1}{2},\frac{1}{2};1,1;4k^2k'^2\right)$

Hence, the desired result follows.

### Re: Series Contest (SC3)

Fri Dec 13, 2013 6:42 pm

Posts: 38
Location: India, West Bengal
Okay, I pass it to anyone who wants to post the next problem.

### Re: Series Contest (SC3)

Thu Dec 19, 2013 2:01 pm

Posts: 852
Location: Jaipur, India

Here is another problem for anybody interested:

Problem 9

$\displaystyle \sum_{n=0}^\infty \frac{1}{F_{2n+1}+1}=\frac{1}{2}\sqrt5$

where $F_n$ is a Fibonacci Number.

### Re: Series Contest (SC3)

Sat Dec 21, 2013 5:20 pm

Posts: 38
Location: India, West Bengal
Am I correct in thinking that this has a whole lot to do with thetas?

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