Is the product $\prod_{n=1}^{\infty}\big(n\beta(n,\frac{1}{2})\big)^{\frac{1}{n}}$ convergent? Consider the following product
$$\prod_{n=1}^{\infty}\big(n\beta(n,\frac{1}{2})\big)^{\frac{1}{n}}$$
where $\beta$ is the Beta function.
I know that the sequence $(n\beta(n,\frac{1}{2})\big)^{\frac{1}{n}}$ is decreasing. Is the product convergent? If yes, does it have any closed form? Thanks.
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\ln\pars{\bracks{n\,\mrm{B}\pars{n,{1 \over 2}}}^{1/n}} & =
{\ln\pars{n} \over n} + {\ln\pars{\mrm{B}\pars{n,1/2}} \over n}
\\[5mm] & =
{\ln\pars{n} + \ln\pars{\Gamma\pars{1/2}} \over n} + {1 \over n}
\ln\pars{\Gamma\pars{n} \over \Gamma\pars{n + 1/2}}
\end{align}

Note that $\ds{\left.\Gamma\pars{n + \alpha}\right\vert_{\ \alpha\ \in\ \mathbb{C}}
\,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\,
\Gamma\pars{n}n^{\alpha}}$.

Then,
\begin{align}
\ln\pars{\bracks{n\,\mrm{B}\pars{n,{1 \over 2}}}^{1/n}} &
\,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\,
{\ln\pars{n} + \ln\pars{\pi}/2 \over n} - {\ln\pars{n} \over 2n}
\,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\,
\color{red}{\ln\pars{n} \over 2n}
\end{align}
A: It diverges. Let's take a logarithm to make product into summation. Then each $n$-th term is $\text{log}\left(n\beta(n,1/2)\right)/n$. If $c_n=\text{log}\left(n\beta(n,1/2)\right)$ has a contant positive lower bound, then the series must diverge (due to harmonic series). Check that $c_n$ is increasing with $c_1= \text{log}2$. (Use definition of beta function, property of gamma function, and relation between $c_n$ and $c_{n-1}$! It is not so hard.)
