Approximation involving Gamma function: $\frac{\Gamma(j+d)}{\Gamma(j+1)\Gamma(d)}\approx(j-1)^{d-1}$

Question

With $d\leq 1$ and $$ a_j=\frac{\Gamma(j+d)}{\Gamma(j+1)\Gamma(d)}=\frac{d(d+1)\ldots(d+j-1)}{j!},\quad j=0,1,2,\ldots $$ my professor wrote in class that $$ \sum_{j=N}^\infty |a_j|\approx\sum_{j=N}^\infty(j-1)^{d-1}\approx\int_N^\infty x^{d-1}dx=\left.\frac{x^d}{d}\right|_N^\infty. $$ (Context: we were mainly concerned with the boundedness of $\sum_j |a_j|$.)

How do I understand/justify the first approximation above? I tried playing with Stirling approximation but I didn't get anything. I have very little experience working with the Gamma function.

Thank you for your help.

Answer 1

Hint: $n~(n-1)~(n-2)~\cdots~(n-k)\approx n^{k+1}$ for large n and fixed k.

Answer 2

First of all, assuming $j,d$ are positive integers, $$ a_j = \frac{\Gamma(j+d)}{\Gamma{j+1}\Gamma{d}} = \frac{(j+d-1)!}{j!(d-1)!} = \frac{(j+d-1)\cdots(j+1)}{(d-1)!}. $$ Using $1+x \leq \exp x$, you can bound $$ \begin{align*} a_j &\leq (j+d-1)\cdots(j+1) \\ &= (j+1)^{d-1} \left(1+\frac{d-2}{j}\right) \left(1+\frac{d-3}{j}\right) \cdots \left(1+\frac{0}{j}\right) \\ &= (j+1)^{d-1} \exp \frac{(d-1)(d-2)}{2j}. \end{align*} $$ If $(d-1)(d-2) \leq 2j$ then $$ a_j \leq e (j+1)^{d-1}. $$ More accurately, we get $$ a_j \leq (j+1)^{d-1} \left(1 + \frac{e}{2} \frac{d^2}{j}\right), $$ so in particular, if $d^2 \ll j$, we get a bound of the form $a_j = (j+1)^{d-1} (1 + o(1))$. A matching lower bound can be obtained using similar methods.