How Can I calculate this Infinite Product?

Question

I would like to calculate the value of this infinite product: $$\prod_{n=1}^\infty \bigg(\frac{n+1}{n}\bigg)^{\frac{1}{n}}$$ According to Wolfram, it seems to converge at around $3.514$, but I have no idea how to start proving this. I have virtually no experience evaluating infinite products.

Any hints? Please don't hand me the entire answer... I would like to try to do some of it on my own.

Answer 1

The given product equals

$$ \exp\sum_{n\geq 1}\frac{1}{n}\log\left(1+\frac{1}{n}\right)=\exp\left[\sum_{n\geq 1}\sum_{m\geq 1}\frac{(-1)^{m+1}}{mn^{m+1}}\right]=\exp\left[\sum_{m\geq 1}\frac{(-1)^{m+1}\zeta(m+1)}{m}\right]$$ that can be rearranged as $$ 2\,\exp\left[\sum_{m\geq 1}\frac{(-1)^{m+1}}{m}\left(\zeta(m+1)-1\right)\right]$$

Thanks to user1952009, by exploiting the integral representation for the $\zeta$ function we have that the product can be written as $$\exp\int_{0}^{+\infty}\frac{\gamma+\Gamma(0,x)+\log(x)}{e^x-1}\,dx\stackrel{\text{IBP}}{=}\exp\left[-\int_{0}^{+\infty}(1-e^{-x})\log(1-e^{-x})\frac{dx}{x}\right]\approx e^{5/4} $$

I doubt there is a "nice" closed form, however.

Answer 2

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

$\ds{\mc{Q}_{N} \equiv \prod_{n = 1}^{N}\pars{n + 1 \over n}^{1/n}\,,\qquad\lim_{N \to \infty}\mc{Q}_{N} =\ {\large ?}}$

\begin{align} \ln\pars{\mc{Q}_{N}} & = \sum_{n = 1}^{N}{1 \over n}\,\ln\pars{n + 1 \over n} = \sum_{n = 1}^{N}{1 \over n}\int_{0}^{1}{\dd t \over n + t} = \int_{0}^{1}\sum_{n = 0}^{N - 1}{1 \over \pars{n + 1}\pars{n + 1 + t}}\dd t \\[5mm] & = \int_{0}^{1}\sum_{n = 0}^{N - 1}\pars{{1 \over n + 1} - {1 \over n + 1 + t}} {\dd t \over t} \\[5mm] & = \int_{0}^{1}\pars{H_{N} -\sum_{n = 0}^{N - 1}{1 \over n + 1 + t}} {\dd t \over t}\qquad\pars{~H_{z}: Harmonic\ Number~} \\[5mm] & = \int_{0}^{1}\bracks{H_{N} - \sum_{n = 0}^{\infty}\pars{{1 \over n + 1 + t} - {1 \over n + N + 1 + t}}} {\dd t \over t} = \int_{0}^{1}{H_{N} - H_{t + N} + H_{t} \over t}\dd t \\[5mm] & \implies \bbx{\lim_{N \to \infty}\ln\pars{\mc{Q}_{N}} = \int_{0}^{1}{H_{t} \over t}\dd t} \end{align}

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$$ \bbox[#ffe,15px,border:1px dotted navy]{\ds{% \prod_{n = 1}^{\infty}\pars{n + 1 \over n}^{1/n} = \exp\pars{\int_{0}^{1}{H_{t} \over t}\dd t}}}\,,\qquad \pars{\substack{\mbox{It seems a 'closed' expression,}\\ \mbox{beyond this one}\\[2mm] \mbox{is}\ {\large highly\ unlikely}}} $$

A 'numerical approach' yields $\ds{\approx 3.5175 \approx \expo{5.031/4}}$.