Consider the infinite product $$\prod_{n=1}^{\infty}\left(\frac{\Gamma(n+\frac{1}{2})}{\sqrt{n}\Gamma(n)}\right).$$ I know that it has the necessary condition for convergence ( i.e., $\frac{\Gamma(n+\frac{1}{2})}{\sqrt{n}\Gamma(n)}\to 1$) and the sequence $\frac{\Gamma(n+\frac{1}{2})}{\sqrt{n}\Gamma(n)}$ is increasing. Is the product convergent? If yes, anyone can determinate its value exactly or represent it by a closed form?

  • $\begingroup$ We have $\frac{\Gamma(n+\frac{1}{2})}{\sqrt{n}\Gamma(n)} = 1 - \frac{1}{8n} + \mathcal{O}\left(\frac{1}{n^2}\right)$ as $n\to\infty$, hence the product $\prod_{n=1}^{N} \frac{\Gamma(n+\frac{1}{2})}{\sqrt{n}\Gamma(n)}$ decreases to zero at the speed of $N^{-1/8}$ as $N\to\infty$. $\endgroup$ – Sangchul Lee Apr 18 '18 at 5:13
  • $\begingroup$ I said that the sequence $\frac{\Gamma(n+\frac{1}{2})}{\sqrt{n}\Gamma(n)}$ is increasing and I am sure for it $\endgroup$ – soodehMehboodi Apr 18 '18 at 5:17
  • $\begingroup$ I am not sure why you are pointing out that fact here. That is not quite relevant to the fate of the infinite product. Indeed, if you have a sequence $a_n$ taking values in $(0, 1)$ (which is true for $a_n = \frac{\Gamma(n+\frac{1}{2})}{\sqrt{n}\Gamma(n)}$), it is always true that their product $p_n = a_1 \cdots a_n$ decreases. $\endgroup$ – Sangchul Lee Apr 18 '18 at 5:20
  • $\begingroup$ I think that perhaps, it is useful to check the convergence of product by some theorems. $\endgroup$ – soodehMehboodi Apr 18 '18 at 5:29
  • $\begingroup$ @soodehMehboodi The sequence $a_n=\frac{n}{n+1}$ is increasing to $1$, but the product of the first $N$ terms of that sequence is $\frac{1}{N+1}$, which goes to $0$. As long as the terms increase (even to $1$) slowly enough, the product can still diverge to $0$. $\endgroup$ – Carl Schildkraut Apr 18 '18 at 5:32

To flesh out Sangchul Lee's comment:

Consider the product of the first $N$ terms:

$$\prod_{n=1}^N \frac{\Gamma(n+1/2)}{\sqrt{n}\Gamma(n)}.$$

As $$\Gamma\left(n+\frac12\right) = \frac{(2n)!}{4^n n!}\sqrt{\pi}$$

and $\Gamma(n)=(n-1)!=\frac{n!}{n},$ we have that this product equals

$$\prod_{n=1}^N \frac{(2n)!\sqrt{n}\sqrt{\pi}}{4^n n! n!}.$$

$$\prod_{n=1}^N \frac{\sqrt{n\pi}}{4^n}\binom{2n}{n}.$$

We may take out the $\sqrt{n\pi}$ and $4^n$ from the product to give

$$\frac{\sqrt{N!} \pi^{N/2}}{2^{N(N+1)}} \prod_{n=1}^N \binom{2n}{n}.$$

By a result given here, this product is asymptotic to

$$\frac{A^{3/2} \cdot 2^{N^2 + N - 7/24} \cdot e^{N/2 - 1/8}}{\pi^{(N+1)/2}\cdot N^{N/2 + 3/8}},$$

where $A$ is the Glaisher-Kinkelin constant. So, the quantity we desire, using Stirling's approximation on $\sqrt{N!}$, is

$$\lim_{N\to\infty} \frac{(2\pi N)^{1/4}\cdot N^{N/2}\cdot \pi^{N/2}\cdot A^{3/2} \cdot 2^{N^2 + N - 7/24} \cdot e^{N/2 - 1/8}}{e^{N/2}\cdot \pi^{(N+1)/2}\cdot N^{N/2 + 3/8}\cdot 2^{N^2+N}}.$$

This simplifies to

$$\frac{A^{3/2}}{e^{1/8} 2^{1/24}\pi^{1/4}}\left(N^{-1/8}\right),$$

so your product diverges to $0$ at the rate of $N^{-1/8}$ with a constant of about $0.93525890...$

  • $\begingroup$ Nice demonstration of asymptotic formula! (+1) $\endgroup$ – Sangchul Lee Apr 18 '18 at 5:33
  • $\begingroup$ @SangchulLee Thanks! Do you know where I could find a better reference for the asymptotic formula than an OEIS comment? I haven't been able to find one, and I don't particularly like just using the comment without any proof behind it. $\endgroup$ – Carl Schildkraut Apr 18 '18 at 5:34
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    $\begingroup$ I am also curious about that. Using the Euler-MacLaurin formula I could check that $$ \prod_{n=1}^{N} \binom{2n}{n} = \frac{2^{N^2+N} e^{\frac{1}{2}N-\frac{1}{8}-\frac{3}{8}\gamma + C}}{\pi^{\frac{1}{2}N} N^{\frac{1}{2}N+\frac{3}{8}}} \left( 1 + \mathcal{O}(N^{-1}) \right), $$ where $$C = \sum_{k=1}^{\infty} (k-1) \left( \log\left(1 - \frac{1}{2k}\right) + \frac{1}{2} - \frac{3}{8k} \right).$$ So it boils down to figuring out the value of $B$, and perhaps it may be computed by utilizing certain zeta functional equation, but I have not pursue this direction yet. $\endgroup$ – Sangchul Lee Apr 18 '18 at 6:46

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