<s>[INST] 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!
 [/INST] 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)$.
</s>