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An interesting infinite product with the closed form,

How can we show show that the formula is correct $$\prod_{n=1}^{\infty}\left(\frac{\Gamma\left(2^n+\frac{1}{2}\right)}{a^n\Gamma(2^n)}\right)^{2^{-n}}=\frac{8}{a^2}\cdot \frac{\sqrt{\pi}}{e^2}$$

Let

$$\prod_{n=1}^{\infty}\left(\frac{\Gamma\left(2^n+\frac{1}{2}\right)}{a^n\Gamma(2^n)}\right)^{2^{-n}}=X$$

take the log

$$\sum_{n=1}^{\infty}\frac{1}{2^n}\log\frac{\Gamma\left(2^n+\frac{1}{2}\right)}{a^n\Gamma(2^n)}=\log X$$

$$\sum_{n=1}^{\infty}\frac{1}{2^n}\log\Gamma\left(2^n+\frac{1}{2}\right)-\sum_{n=1}^{\infty}\frac{1}{2^n}\log\Gamma\left(2^n\right)-\sum_{n=1}^{\infty}\frac{n}{2^n}\log a=\log X$$

$$\sum_{n=1}^{\infty}\frac{1}{2^n}\log\Gamma\left(2^n+\frac{1}{2}\right)-\sum_{n=1}^{\infty}\frac{1}{2^n}\log\Gamma\left(2^n\right)-\log a^2=\log X$$

$$\sum_{n=1}^{\infty}\frac{1}{2^n}\log\Gamma\left(2^n+\frac{1}{2}\right)-\sum_{n=1}^{\infty}\frac{1}{2^n}\log\Gamma\left(2^n\right)=\log(a^2X)$$

$$\sum_{n=1}^{\infty}\frac{1}{2^n}\log\Gamma\left(2^n+\frac{1}{2}\right)-\sum_{n=1}^{\infty}\frac{\log(2^n-1)!}{2^n}=\log(a^2X)$$

...

Ron Gordon
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Sibawayh
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    Could try using this: ${\displaystyle \Gamma (z)\Gamma \left(z+{\tfrac {1}{2}}\right)=2^{1-2z};{\sqrt {\pi }};\Gamma (2z).}$ – fGDu94 Dec 23 '19 at 13:10
  • Generate the finite product and take the limit as the number of terms goes to infinity. – Ron Gordon Dec 23 '19 at 14:02
  • Showing that $\prod_{n=1}^\infty\frac{1}{a^{n/2^n}}=1/a^2$ comes down to the fact that $\sum_{n=1}^\infty n/2^n=2$. The rest of the problem is done here. – dxdydz Dec 23 '19 at 17:10

1 Answers1

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As I said in the comments, evaluate the finite product and then take the limit as the number of terms, $N \to \infty$.

First let's state up front that

$$\frac{\Gamma{\left ( n+\frac12 \right )}}{\Gamma{(n)}} = \sqrt{\pi} n \frac{(2 n)!}{2^{2 n} (n!)^2} $$

so that we can write the product out to $N$ terms as follows:

$$\left (\sqrt{\pi} \left ( \frac{2}{a^2} \right)^1 \frac{\left (2^{2} \right )!}{2^{2^{2}} \left ( 2^1 \right )!^2} \right )^{1/2^1} \left (\sqrt{\pi} \left ( \frac{2}{a^2} \right)^2 \frac{\left (2^{3} \right )!}{2^{2^{3}} \left ( 2^2 \right )!^2} \right )^{1/2^2} \cdots \left (\sqrt{\pi} \left ( \frac{2}{a^2} \right)^N \frac{\left (2^{N+1} \right )!}{2^{2^{N+1}} \left ( 2^N \right )!^2} \right )^{1/2^N}$$

which can be simplified to a single term as follows. First, the power of the $\sqrt{\pi}$ term is a geometric series: $$\frac12 + \frac1{2^2} + \cdots +\frac1{2^N} = 1-\frac1{2^N}$$

Then the power of the $2/a^2$ term is another well-known series:

$$\frac12 + \frac{2}{2^2} + \cdots +\frac{N}{2^N} = 2-\frac{N+2}{2^N}$$

Finally, we get cancellations of the $\left ( 2^{n+1} \right )!$ terms in the denominator, leaving one of the corresponding $\left ( 2^{n} \right )!$ in the denominator. The product is then, exactly,

$$\left ( \sqrt{\pi} \right )^{1-2^{-N}} \left (\frac{2}{a^2} \right )^{2-(N+2) 2^{-N}} 2^{-(2 N+1)} \left ( \left ( 2^{N+1} \right )! \right )^{2^{-N}}$$

Now we consider the limit as $N \to \infty$. Here we use Stirling and find that

$$\begin{align} \left ( \left ( 2^{N+1} \right )! \right )^{2^{-N}} &\approx (2 \pi)^{2^{-(N+1)}} 2^{2 N+2} 2^{(N+1) 2^{-(N+1)}} e^{-2} \end{align} $$

so that, when plugged into the exact expression, and the limit as $N \to \infty$ is taken, produces as the product $8 \sqrt{\pi}/e^2 1/a^2$, as asserted.

Ron Gordon
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