Is the product $\prod_{n=1}^{\infty}\big(n\beta(n,\frac{1}{2})\big)^{\frac{1}{n}}$ convergent?

Question

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.

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}

Answer 2

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.)