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?
On
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.
On
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.
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\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}
\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}
$$ \bbx{\prod_{n = 1}^{\infty}{\pars{1 + 1/n}^{n} \over \bracks{1 + 1/\pars{n + x}}^{n + x}} = {\expo{-x}\pars{1 + x}^{x} \over x!}} $$