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

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.

• Does it help to know that $\Gamma(n) = (n-1)!$ ? – Danny W. Sep 28 '14 at 22:55
• @ Danny: I know that one :) (I edited the question to reflect that). If you know the explanation to my question, please elaborate. I would like to learn it from you. Thank you! – Kim Jong Un Sep 28 '14 at 22:56
• I may not have seen the part $d \leq 1$ - this makes it a little more difficult, since you then can't necessarily use the factorial function. – Danny W. Sep 28 '14 at 22:57

Hint: $n~(n-1)~(n-2)~\cdots~(n-k)\approx n^{k+1}$ for large n and fixed k.
• @ Lucian: forgive my obtuseness, but I think in this case, my $k$ isn't fixed. – Kim Jong Un Sep 28 '14 at 23:11
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.
• @ Yuval: thank you very much and +1. I edited the question slightly to change $a_j$ to $|a_j|$. I apologize, but could you please adjust to this change? Your solution will be very instructive to me. – Kim Jong Un Sep 28 '14 at 23:34
• Assuming $j+\lceil d \rceil \geq 1$, you can use $\Gamma(j+d) \leq \Gamma(j+ \lceil d \rceil)$ together with the recurrence $\Gamma(j+1) = j(j-1)\cdots(j+\lceil d\rceil) \Gamma(j+\lceil d\rceil)$ to obtain similar results. – Yuval Filmus Sep 28 '14 at 23:39