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## Infinite product

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### Infinite product

Thu Jul 21, 2016 12:20 pm

Posts: 48

Prove that:

$$\prod_{m=0}^{\infty} \frac{(4m+3)(4m+3)(4m+6)}{(4m+2)(4m+5)(4m+5)}=\frac{1}{16\pi^2}\,\Gamma^4\left(\frac{1}{4}\right)$$

or equivelantly prove that:

$$\sum_{n=1}^{\infty}\left[\frac{\lambda(2n)}{n}-\ln\left(\frac{n+1}{n}\right)\right] =\gamma+\ln\left(\frac{16\pi^2}{\Gamma^4\left(\frac{1}{4}\right)}\right)$$

where $\lambda(s)=\sum \limits_{n=0}^{\infty} \frac{1}{(2n+1)^s}$.

### Re: Infinite product

Thu Aug 04, 2016 10:50 pm

Posts: 5
VJKey wrote:
Prove that:
$$\prod_{m=0}^{\infty} \frac{(4m+3)(4m+3)(4m+6)}{(4m+2)(4m+5)(4m+5)}=\frac{1}{16\pi^2}\,\Gamma^4\left(\frac{1}{4}\right)$$

Is it well written the term (4k+3)*(4k+3) ?
Last edited by pprime on Fri Aug 05, 2016 2:46 pm, edited 1 time in total.

### Re: Infinite product

Thu Aug 04, 2016 11:54 pm

Posts: 5
$\displaystyle P=\prod\limits_{m=0}^{+\infty }{\frac{\left( 4m+3 \right)\left( 4m+3 \right)\left( 4m+6 \right)}{\left( 4m+2 \right)\left( 4m+5 \right)\left( 4m+5 \right)}}=\frac{27}{25}\prod\limits_{m=1}^{+\infty }{\frac{\left( 1+\frac{3}{4m} \right)\left( 1+\frac{3}{4m} \right)\left( 1+\frac{6}{4m} \right)}{\left( 1+\frac{2}{4m} \right)\left( 1+\frac{5}{4m} \right)\left( 1+\frac{5}{4m} \right)}}$

$\displaystyle =\frac{27}{25}\prod\limits_{m=1}^{+\infty }{\frac{\left( 1+\frac{2}{4m} \right)^{-1}\left( 1+\frac{5}{4m} \right)^{-1}\left( 1+\frac{5}{4m} \right)^{-1}}{\left( 1+\frac{3}{4m} \right)^{-1}\left( 1+\frac{3}{4m} \right)^{-1}\left( 1+\frac{6}{4m} \right)^{-1}}}$

$\displaystyle =\frac{27}{25}\frac{\prod\limits_{m=1}^{+\infty }{\left( 1+\frac{2}{4m} \right)^{-1}e^{\frac{2}{4m}}\cdot }\prod\limits_{m=1}^{+\infty }{\left( 1+\frac{5}{4m} \right)^{-1}e^{\frac{5}{4m}}\cdot }\prod\limits_{m=1}^{+\infty }{\left( 1+\frac{5}{4m} \right)^{-1}e^{\frac{5}{4m}}}}{\prod\limits_{m=1}^{+\infty }{\left( 1+\frac{3}{4m} \right)^{-1}e^{\frac{3}{4m}}\cdot }\prod\limits_{m=1}^{+\infty }{\left( 1+\frac{3}{4m} \right)^{-1}e^{\frac{3}{4m}}}\cdot \prod\limits_{m=1}^{+\infty }{\left( 1+\frac{6}{4m} \right)^{-1}e^{\frac{6}{4m}}}}$

$\displaystyle =\frac{\left( \frac{4}{2}e^{-\frac{2}{4}\gamma }\prod\limits_{m=1}^{+\infty }{\left( 1+\frac{2}{4m} \right)^{-1}e^{\frac{2}{4m}}} \right)\cdot \left( \frac{4}{5}e^{-\frac{5}{4}\gamma }\prod\limits_{m=1}^{+\infty }{\left( 1+\frac{5}{4m} \right)^{-1}e^{\frac{5}{4m}}} \right)\cdot \left( \frac{4}{5}e^{-\frac{5}{4}\gamma }\prod\limits_{m=1}^{+\infty }{\left( 1+\frac{5}{4m} \right)^{-1}e^{\frac{5}{4m}}} \right)}{\left( \frac{4}{3}e^{-\frac{3}{4}\gamma }\prod\limits_{m=1}^{+\infty }{\left( 1+\frac{3}{4m} \right)^{-1}e^{\frac{3}{4m}}} \right)\cdot \left( \frac{4}{3}e^{-\frac{3}{4}\gamma }\prod\limits_{m=1}^{+\infty }{\left( 1+\frac{3}{4m} \right)^{-1}e^{\frac{3}{4m}}} \right)\cdot \left( \frac{4}{6}e^{-\frac{6}{4}\gamma }\prod\limits_{m=1}^{+\infty }{\left( 1+\frac{6}{4m} \right)^{-1}e^{\frac{6}{4m}}} \right)}$

$\displaystyle =\frac{\Gamma \left( \frac{2}{4} \right)\cdot \Gamma \left( \frac{5}{4} \right)\cdot \Gamma \left( \frac{5}{4} \right)}{\Gamma \left( \frac{3}{4} \right)\cdot \Gamma \left( \frac{3}{4} \right)\cdot \Gamma \left( \frac{6}{4} \right)}=\frac{\sqrt{\pi }\cdot \Gamma ^{2}\left( \frac{5}{4} \right)}{\Gamma \left( \frac{3}{2} \right)\cdot \Gamma ^{2}\left( \frac{3}{4} \right)}=\frac{\sqrt{\pi }\cdot \left( \frac{\Gamma \left( \frac{1}{4} \right)}{4} \right)^{2}}{\frac{\sqrt{\pi }}{2}\cdot \left( \frac{\sqrt{2}\pi }{\Gamma \left( \frac{1}{4} \right)} \right)^{2}}=\frac{\Gamma ^{3}\left( \frac{1}{4} \right)}{16\pi ^{2}}$