Equivalent of $ x(x+1)(x+2)\cdots(x+n)$? 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!
 A: 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.
A: 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.
A: 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)$.
