Home » Uncategorized » Gamma infinite product

Gamma infinite product

Prove that

    \[\prod_{n=1}^\infty\sqrt[2^n]{\frac{\Gamma(2^n+\frac{1}{2})}{\Gamma(2^n)}} = \frac{8\sqrt{\pi}}{e^2}\]

Solution

Converting the product to a sum and using duplication formula for the gamma function and telescoping,

\displaystyle \sum_{n=1}^{N}\frac{1}{2^n}\ln\frac{\Gamma(2^n+\frac12)}{\Gamma(2^n)}=\left(1-2^{-N}\right)\ln\left(2\sqrt{\pi}\right)-2N\ln2+\frac{2}{2^{N+1}}\ln\Gamma(2^{N+1})

Using Stirling formula

    \[\frac{1}{N}\ln\Gamma(N)=\ln N-1+\mathcal{O}\left(\frac{\ln N}{N}\right)\quad \text{as}\;\; N\rightarrow\infty\]

we get that

    \[\sum_{n=1}^{\infty}\frac{1}{2^n}\ln\frac{\Gamma(2^n+\frac12)}{\Gamma(2^n)}=\ln\left(2\sqrt{\pi}\right)+2\left(\ln 2-1\right)=\ln\frac{8\sqrt{\pi}}{e^2}\]

Read more

Leave a comment

Who is Tolaso?

Find out more at his Encyclopedia Page.

Donate to Tolaso Network