How Can I calculate this Infinite Product? 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. 
 A: 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.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\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}

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


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

