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Is it true that $$\displaystyle\prod_{n=1}^\infty\frac{(4n-3)(4n+1)}{(4n-1)^2}=\frac{8\pi^2}{\left(\Gamma\left(\frac{1}{4}\right)\right)^4}?$$

How to prove it ? Thank in advances.

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3 Answers 3

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Yes. using the the Gamma function in the form $$\Gamma\left(z\right)=\lim_{n\rightarrow\infty}\frac{n^{z-1}n!}{z\left(z+1\right)\cdots\left(z+n-1\right)}$$ we can prove that $$\prod_{n\geq0}\frac{\left(n+a\right)\left(n+b\right)}{\left(n+c\right)\left(n+d\right)}=\frac{\Gamma\left(c\right)\Gamma\left(d\right)}{\Gamma\left(a\right)\Gamma\left(b\right)},\, a+b=c+d $$ so $$\prod_{n\geq1}\frac{\left(4n-3\right)\left(4n+1\right)}{\left(4n-1\right)\left(4n-1\right)}=\prod_{n\geq0}\frac{\left(n+1/4\right)\left(n+5/4\right)}{\left(n+3/4\right)\left(n+3/4\right)}=\frac{\Gamma\left(\frac{3}{4}\right)^{2}}{\Gamma\left(\frac{1}{4}\right)\Gamma\left(\frac{5}{4}\right)}=\color{red}{\frac{8\pi^{2}}{\Gamma\left(\frac{1}{4}\right)^{4}}}. $$

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  • $\begingroup$ Thank you. So we can conclude that $\sum_{n=1}^\infty (-1)^{n+1}\ln{\left(\frac{2n-1}{2n+1}\right)}=\ln{\left(\frac{8\pi^2}{\Gamma\left(\frac{1}{4}\right)^4}\right)}$ . ? $\endgroup$
    – rack
    Commented Dec 9, 2016 at 17:52
  • $\begingroup$ @rack Yes, we can. $\endgroup$ Commented Dec 9, 2016 at 17:55
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$$ \begin{align} \prod_{n=1}^\infty\frac{(4n-3)(4n+1)}{(4n-1)^2} &=\lim_{m\to\infty}\prod_{n=1}^m\frac{\left(n-\frac34\right)\left(n+\frac14\right)}{\left(n-\frac14\right)^2}\\ &=\lim_{m\to\infty}\frac{\left.\Gamma\left(m+\frac14\right)\middle/\Gamma\left(\frac14\right)\right.\,\,\left.\Gamma\left(m+\frac54\right)\middle/\Gamma\left(\frac54\right)\right.}{\left(\left.\Gamma\left(m+\frac34\right)\middle/\Gamma\left(\frac34\right)\right.\right)^2}\\ &=\frac{\Gamma\left(\frac34\right)^2}{\Gamma\left(\frac14\right)\Gamma\left(\frac54\right)} \lim_{m\to\infty}\frac{\Gamma\left(m+\frac14\right)m^{1/2}}{\Gamma\left(m+\frac34\right)} \lim_{m\to\infty}\frac{\Gamma\left(m+\frac54\right)m^{-1/2}}{\Gamma\left(m+\frac34\right)}\\ &=4\frac{\left(\Gamma\left(\frac14\right)\Gamma\left(\frac34\right)\right)^2}{\Gamma\left(\frac14\right)^4}\\ &=\frac{8\pi^2}{\Gamma\left(\frac14\right)^4} \end{align} $$ Gautschi's Inequality says the last two limit are $1$. Euler's Reflection Formula, a proof of which is given at the end of this answer, says that $\Gamma\left(\frac14\right)\Gamma\left(\frac34\right)=\pi\csc(\pi/4)=\pi\sqrt2$.

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  • $\begingroup$ Thank you. So we can conclude that $\sum_{n=1}^\infty (-1)^{n+1}\ln{\left(\frac{2n-1}{2n+1}\right)}=\ln{\left(\frac{8\pi^2}{\Gamma\left(\frac{1}{4}\right)^4}\right)}$ . ? $\endgroup$
    – rack
    Commented Dec 9, 2016 at 17:53
  • $\begingroup$ That looks right $\endgroup$
    – robjohn
    Commented Dec 9, 2016 at 17:54
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The relation is true.

The hard part is to demonstrate that $$ \prod_{n=1}^\infty\frac{(4n-3)(4n+1)}{(4n-1)^2}=\frac{4\left(\Gamma\left(\frac34\right)\right)^2}{\left(\Gamma\left(\frac14\right)\right)^2} $$

Once you have that, you can use the reflection property which says $$ \Gamma\left(\frac34\right)\Gamma\left(\frac14\right)= \frac{\pi}{\sin(\pi/4)} = \sqrt{2}\pi $$ and the result becomes simple multiplication.

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