Is the product $\prod_{n=1}^{\infty}\left(\frac{\Gamma(n+\frac{1}{2})}{\sqrt{n}\Gamma(n)}\right)$ convergent? 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? 
 A: 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 $0.93525890...$
