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

Solution 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.

Solution 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.