Equivalent of $ x(x+1)(x+2)\cdots(x+n)$?

Question

Assume $x>0$. Is there an equivalent for this quantity $$ x(x+1)(x+2)\cdots(x+n)$$ as $n$ tends to $+\infty$?

I've tried to write $$x(x+1)(x+2)\cdots(x+n)=x^{n+1}\left(1+\frac 1x\right)\left(1+\frac 2x\right)\cdots\left(1+\frac nx\right)$$ I don't know if I'm on the right track... Thanks for your help!

Answer 1

Hint. If $x$ was $1$, this would be asking for an equivalent of $n!$, so you'd need Stirling's formula. In general, the function you've written is equal to $\frac{1}{\Gamma(x)} \Gamma(x + n + 1)$, so you can still obtain an equivalent directly from Stirling's formula, which is also applicable to the Gamma function.

Answer 2

Let $x$ be a real number such that $x>0$.

One may observe that, using successive integrations by parts, we have $$ \int_0^n t^{x-1} \left( 1-\frac{t}{n}\right)^n{\rm{d}} t= \frac{n! \:n^x}{x(x+1)(x+2)\cdots(x+n)},\quad n=1,2,\ldots, $$ leading to $$ \Gamma(x)=\lim_{n\to+\infty}\left(\frac{n! \:n^x}{x(x+1)(x+2)\cdots(x+n)}\right). $$ Then, as $n$ is great, the desired equivalent is $$ x(x+1)(x+2)\cdots(x+n) \sim \frac{n! \:n^x}{\Gamma(x)} $$ or $$ x(x+1)(x+2)\cdots(x+n) \sim \frac{n^{n+x}e^{-n}\sqrt{2\pi n} }{\Gamma(x)} $$ with Stirling's formula.

Answer 3

I showed many years ago that the closest $n$-th power of an integer to $x(x+1)...(x+n-1)$ for integer $x$ is $(x+[(n-1)/2])^n$ whenever $x$ is large enough compared to $n$.

A sufficient condition is, with $u = x+[(n-1)/2]$, $u \geq (n^2-n)/4$ for even $n$ and $u \geq (n^2+3n-4)/4$ for odd $n$.

Some more precise results, with $u = x+[(n-1)/2]$ (the proofs of which are not easy):

The smallest value of $u$ such that $u^n$ is closer to $x(x+1)...(x+n-1)$ than $(u-1)^n$ is $ \dfrac{n^2-1}{12} + \dfrac{n-1}{4} + \dfrac{13}{20} + O\left(\dfrac1{n}\right)$.

The smallest value of $u$ such that $(u-1/2)^n$ is closer to $x(x+1_...(x+n-1)$ than $(u-3/2)^n$ is $ \dfrac{n^2-1}{24} + \dfrac{n-1}{8} + \dfrac{13}{10} + O\left(\dfrac1{n}\right)$.