Find the Limit of $‎\prod‎_{n=1}^{‎\infty}\frac{(1+‎\frac{1}{n}‎)^n}{(1+‎\frac{1}{n+x})^{n+x}}$ ‎Consider the following productions‎
‎$$‎‎‎‎‎\prod‎_{n=1}^{‎\infty}\frac{1+‎\frac{1}{n}‎}{1+‎\frac{1}{n+x}}$$
and
$$‎‎‎‎‎\prod‎_{n=1}^{‎\infty}\frac{(1+‎\frac{1}{n}‎)^n}{(1+‎\frac{1}{n+x})^{n+x}}$$
I know that the above are convergent.
Can anyone find the limit of these products?
‎
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\Large\left. a\right)}$

\begin{align}
&\bbox[10px,#ffd]{\prod‎_{n = 1}^{‎N}{1 + 1/n \over
1 + 1/\pars{n+x}}} =
{\prod‎_{n = 1}^{‎N}\pars{n + 1} \over
\prod‎_{n = 1}^{‎N}n}\,
{\prod‎_{n = 1}^{‎N}\pars{n + x} \over
\prod‎_{n = 1}^{‎N}\pars{n + x + 1}}
\\[5mm] = &\
{\prod‎_{n = 2}^{‎N + 1}n \over \prod‎_{n = 1}^{‎N}n}\,
{\prod‎_{n = 1}^{‎N}\pars{n + x} \over \prod‎_{n = 2}^{‎N + 1}\pars{n + x}}
\\[5mm] = &\
\pars{N + 1}\,
{\pars{1 + x}\prod‎_{n = 2}^{‎N}\pars{n + x} \over
\pars{N + 1 + x}\prod‎_{n = 2}^{‎N}\pars{n + x}}
\,\,\,\stackrel{\mrm{as}\ N\ \to\ \infty}{\to}\,\,\,\bbx{1 + x}
\end{align}

$\ds{\Large\left. b\right)}$
$\ds{\prod‎_{n = 1}^{‎\infty}{\pars{1 + 1/n‎}^{n} \over
\bracks{1 + ‎1/\pars{n + x}}^{n + x}}:\ {\Large ?}}$.

\begin{align}
&\bbox[10px,#ffd]{\prod_{n = 1}^{N}
\pars{1 + {1 \over n + a}}^{n + a}} =
{\prod_{n = 1}^{N}\pars{n + 1 + a}^{n + a} \over
\prod_{n = 1}^{N}\pars{n + a}^{n + a}}
\\[5mm] = &\
{\prod_{n = 2}^{N + 1}\pars{n + a}^{n - 1 + a} \over
\prod_{n = 1}^{N}\pars{n + a}^{n + a}}
\\[5mm] = &\
{1 \over \prod_{n = 2}^{N + 1}\pars{n + a}}\,
{\pars{N + 1 + a}^{N + 1 + a}\prod_{n = 2}^{N}
\pars{n + a}^{n  + a} \over
\pars{1 + a}^{1 + a}\prod_{n = 2}^{N}
\pars{n + a}^{n + a}}
\\[5mm] = &\
{1 \over \pars{2 + a}^{\overline{N}}}\,
{N^{N + 1 + a} \over \pars{1 + a}^{1 + a}}\,
\pars{1 + {1 + a \over N}}^{N + 1 + a}
\\[5mm] = &\
{\Gamma\pars{2 + a} \over \Gamma\pars{2 + a + N}}\,
{N^{N + 1 + a} \over \pars{1 + a}^{1 + a}}\,
\pars{1 + {1 + a \over N}}^{N + 1 + a}
\\[5mm] \stackrel{\mrm{as}\ N\ \to\ \infty}{\sim}\,\,\,&
{\Gamma\pars{2 + a}\expo{1 + a} \over
\pars{1 + a}^{1 + a}}\,
{N^{N + 1 + a} \over \Gamma\pars{2 + a + N}}
\end{align}
Then,
\begin{align}
&\bbox[10px,#ffd]{\prod‎_{n = 1}^{‎N}{\pars{1 + 1/n‎}^{n} \over
\bracks{1 + ‎1/\pars{n + x}}^{n + x}}}
\\[5mm] \stackrel{\mrm{as}\ N\ \to\ \infty}{\sim}\,\,\,&
{\Gamma\pars{2}\expo{}N^{N + 1}/\Gamma\pars{2 + N} \over
\Gamma\pars{2 + x}\expo{1 + x}N^{N + 1 + x}/
\bracks{\pars{1 + x}^{1 + x}\,\Gamma\pars{2 + x + N}}}
\\[5mm] = &\
{\expo{-x}\pars{1 + x}^{1 + x} \over
\Gamma\pars{2 + x}}\,
{\pars{N + 1 + x}! \over N^{x}\pars{N + 1}!}
\\[5mm] \stackrel{\mrm{as}\ N\ \to\ \infty}{\sim}\,\,\,&
{\expo{-x}\pars{1 + x}^{1 + x} \over
\Gamma\pars{2 + x}}\,
{\root{2\pi}\pars{N + 1 + x}^{N + 3/2 + x}
\expo{-\pars{N + 1 + x}} \over N^{x}\bracks{\root{2\pi}\pars{N + 1}^{N + 3/2}
\expo{-\pars{N + 1}}}}
\\[5mm] = &\
{\expo{-x}\pars{1 + x}^{1 + x} \over
\Gamma\pars{2 + x}}\,
{N^{N + 3/2 + x}\bracks{1 + \pars{1 + x}/N}^{N + 3/2 + x} \over N^{x}
\bracks{N^{N + 3/2}\,\pars{1 + 1/N}^{N + 3/2}}}
\,\expo{-x}
\\[5mm] \stackrel{\mrm{as}\ N\ \to\ \infty}{\sim}\,\,\,&
{\expo{-x}\pars{1 + x}^{1 + x} \over
\pars{1 + x}\Gamma\pars{1 + x}}\,
{\expo{1 + x} \over \expo{}}\,\expo{-x}
\end{align}

$$
\bbx{\prod‎_{n = 1}^{‎\infty}{\pars{1 + 1/n‎}^{n} \over
\bracks{1 + ‎1/\pars{n + x}}^{n + x}} =
{\expo{-x}\pars{1 + x}^{x} \over x!}}
$$
A: For the first one, rewrite the product: 
\begin{equation*}
\begin{split}
\prod_{n=1}^{\infty} \frac{(n+1)(n+x)}{n(n+x+1)}& = \frac{(1+1)(1+x)}{1(1+x+1)} \times\frac{(2+1)(2+x)}{2(2+x+1)} \times \frac{(3+1)(3+x)}{3(3+x+1)}\times ...\\
& = \frac{2(1+x)}{1(2+x)} \times\frac{3(2+x)}{2(3+x)} \times \frac{4(3+x)}{3(4+x)}\times ...\\
& = (1+x).
\end{split}
\end{equation*}
For the second one, rewrite the product as
\begin{equation*}
\begin{split}
\prod_{n=1}^{\infty} \frac{(n+1)^n(n+x)^{n+x}}{n^n(n+x+1)^{n+x}} & = \frac{(1+1)^1(1+x)^{1+x}}{1^1(1+x+1)^{1+x}} \times \frac{(2+1)^2(2+x)^{2+x}}{2^2(2+x+1)^{2+x}}\times \frac{(3+1)^3(3+x)^{3+x}}{3^3(3+x+1)^{3+x}}\times ...\\
& = \frac{2(1+x)^{1+x}}{(2+x)^{1+x}} \times \frac{3^2(2+x)^{2+x}}{2^2(3+x)^{2+x}}\times \frac{4^3(3+x)^{3+x}}{3^3(4+x)^{3+x}}\times ...\\
& = (1+x)^{1+x} \times \frac{(2+x)}{2}\times \frac{(3+x)}{3}\times ...\\
& = (1+x)^{1+x}\prod_{n=2}^{\infty}\frac{n+x}{n}.
\end{split}
\end{equation*}
The product is equal to zero for $x<0$, it is equal to $1$ for $x=0$, and for $x>0$ the product is divergent.
A: Note that for $x = k \in N$
$$‎‎‎‎‎
\prod‎_{n=1}^{‎\infty}\frac{(1+‎\frac{1}{n}‎)^n}{(1+‎\frac{1}{n+x})^{n+x}}=\frac{1}{e}\prod‎_{n=1}^{‎k}\left(1+‎\frac{1}{n}\right)^n
$$
by cancellation. So for $x > 0$ we have $\prod‎_{n=1}^{‎\infty}\frac{(1+‎\frac{1}{n}‎)^n}{(1+‎\frac{1}{n+x})^{n+x}}\le M(x)$
This result can be extended with the contribution of $\Gamma$ functions.
