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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.

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  • $\begingroup$ Does it help to know that $\Gamma(n) = (n-1)!$ ? $\endgroup$ – Danny W. Sep 28 '14 at 22:55
  • $\begingroup$ @ 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! $\endgroup$ – Kim Jong Un Sep 28 '14 at 22:56
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    $\begingroup$ 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. $\endgroup$ – Danny W. Sep 28 '14 at 22:57
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Hint: $n~(n-1)~(n-2)~\cdots~(n-k)\approx n^{k+1}$ for large n and fixed k.

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  • $\begingroup$ @ Lucian: forgive my obtuseness, but I think in this case, my $k$ isn't fixed. $\endgroup$ – Kim Jong Un Sep 28 '14 at 23:11
  • $\begingroup$ @KimJongUn: As far as I can see, your d doesn't vary, your j does, which is precisely what your teacher used in his first series approximation. Unfortunately, you chose the unhelpful expression, with both varying numerator and varying denominator. $\endgroup$ – Lucian Sep 28 '14 at 23:46
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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.

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  • $\begingroup$ @ 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. $\endgroup$ – Kim Jong Un Sep 28 '14 at 23:34
  • $\begingroup$ 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. $\endgroup$ – Yuval Filmus Sep 28 '14 at 23:39
  • $\begingroup$ @ Yuval: thank you. I will study your answer and will accept after I understand it fully. Meanwhile, if you want to add any details, please feel free to. Regards. $\endgroup$ – Kim Jong Un Sep 28 '14 at 23:41

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