Evaluate product $\prod_{i=1}^{n}(\frac{i}{i+x})^{i}$

Question

Is there a way to evaluate this product so that the answer is in closed form?

$$\prod_{i=1}^{n}\left(\frac{i}{i+x}\right)^{i}=\left(\frac{1}{1+x}\right)\left(\frac{2}{2+x}\right)^{2}\cdots\left(\frac{n}{n+x}\right)^{n}$$

It can of course be written as

$$\prod_{i=1}^{n}\left(\frac{1}{1+\frac{x}{i}}\right)^{i}$$ but I don't know if that helps.

Thank you!

Answer 1

A comment.

$\displaystyle Q_0(x) :=\Gamma(x+1)=\lim_{n\to\infty}\frac{n^x}{\prod\limits_{k=1}^n\left(1+\frac{x}{k}\right)}~~ , ~~~~ Q_1(x) :=\lim_{n\to\infty}\frac{e^{xn}n^{-x^2/2}}{\prod\limits_{k=1}^n\left(1+\frac{x}{k}\right)^k}$

$$\prod\limits_{i=1}^n\left(\frac{i}{i+x}\right)^i=\frac{Q_1(x)Q_1(n)Q_0(n)^n}{Q_1(x+n)Q_0(x+n)^n e^{xn}}$$

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Note: $~$ Closely related to $~Q_1(x)~$ is the $\,$Barnes_G-function $~G(x)$.

$\hspace{1.3cm} Q_1(x)\cdot G(x+1) = Q_1(1)^x\,e^{-\frac{1}{2}x(x-1)}~$ with $~Q_1(1)=\frac{\sqrt{2\pi}}{e}$

$\hspace{1.3cm}$ It follows

$$\prod\limits_{i=1}^n\left(\frac{i}{i+x}\right)^i=\frac{G(x+n+1)}{G(x+1)\,G(n+1)}\left(\frac{\Gamma(n+1)}{\Gamma(x+n+1)}\right)^n$$

$\hspace{1.3cm}$ which can also be inferred from $~\prod\limits_{i=1}^n (i+x)^i = \frac{G(x+1)\,\Gamma(x+n+1)^n}{G(x+n+1)}~ , ~G(1)=1$ .

Answer 2

We start with a little intro to the Hurwitz zeta function: $$\zeta(s,q)=\sum_{k\ge0}\frac{1}{(q+k)^s}.$$ The above is the definition of this function for $\text{Re }s>1$ and $\text{Re }q>0$, although one can extend it to a meromorphic function for $s\in\Bbb C\setminus \{1\}$ and $q\in\Bbb C$. We will be dealing a lot with the function $$\zeta^{(1,0)}(s,q)=\frac{\partial}{\partial s}\zeta(s,q),$$ especially at values of $s$ such as $0$ and $-1$. Our problems with convergence with these cases of $s$ are fixed by the meromorphic extension.

First, if we set $$f_M(s,q)=\sum_{k=0}^{M}\frac{1}{(q+k)^s},$$ it is fairly simple to show that $$f_M(s,q)=\zeta(s,q)-\zeta(s,q+M+1).$$ Anyway, we see that $$f_M^{(1,0)}(s,q)=-\sum_{k=0}^M \frac{\ln(q+k)}{(q+k)^s}.$$ Therefore $$\prod_{k=0}^{M}(q+k)^{q+k}=\exp\left(-f_M^{(1,0)}(-1,q)\right)=\exp\left(\zeta^{(1,0)}(-1,q+M+1)-\zeta^{(1,0)}(-1,q)\right),$$ because $$\sum_{k=0}^{M}a_k \ln b_k=\sum_{k=0}^{M}\ln(b_k^{a_k})=\ln\left[\prod_{k=0}^{M}b_k^{a_k}\right].$$

Next, we see that $$P_n(x)=\prod_{k=1}^{n}\left(\frac{k}{x+k}\right)^k=\frac{H(n)}{q_n(x)}$$ where $H(n)=\prod_{k=1}^{n}k^k$ and $q_n(x)=\prod_{k=1}^{n}(x+k)^{k}$.

Then we see that $$\begin{align} \prod_{k=0}^{n}(x+k)^{x+k}&=x^x\prod_{k=1}^{n}(x+k)^{x+k}\\ &=x^x\left(\prod_{k=1}^{n}(x+k)^x\right)\left(\prod_{k=1}^{n}(x+k)^{k}\right)\\ &=\left(\prod_{k=0}^{n}(x+k)\right)^xq_n(x)\\ &=\left(\frac{\Gamma(x+n+1)}{\Gamma(x)}\right)^xq_n(x). \end{align}$$ In the last step, we used the Gamma function $\Gamma(z)$. For information on the ratio of the two Gamma functions above, see here.

Anyway, we have that $$P_n(x)=\left(\frac{\Gamma(x+n+1)}{\Gamma(x)}\right)^xH(n)\exp\left(\zeta^{(1,0)}(-1,x)-\zeta^{(1,0)}(-1,x+n+1)\right).$$ This is a somewhat 'nicer' version of what I left in the comments.