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I have the following to start:

$$F(x)=x^{b-a}\frac{\Gamma(x+a+1)}{\Gamma(x+b+1)}$$

And I'm trying to evaluate:

$$\lim_{x\rightarrow\infty}F(x)$$

I have simplified this to yield the same outcome as

$$e^{b-a} \lim_{x\rightarrow\infty}x^{b-a}\frac{(x+a)^{x+a}}{(x+b)^{x+b}}$$

But I am totally stuck from here.

Any input on how to evaluate the initial limit or how to evaluate the limit I have simplified to would be excellent. I know that using Stirling's approximation is useful in finding the limit of the initial problem and that is in fact how I reached my simplification.

Thanks in advance.

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  • $\begingroup$ Have you tried using L'Hopital on that limit you obtained? $\endgroup$ – Prasun Biswas Apr 8 '15 at 7:01
  • $\begingroup$ @PrasunBiswas I just corrected the limit I obtained because I made a mistake and left out part. Does this thought still apply? $\endgroup$ – Nick Chapman Apr 8 '15 at 7:03
  • $\begingroup$ Is $b\geq a$ in the question? $\endgroup$ – Prasun Biswas Apr 8 '15 at 7:05
  • $\begingroup$ @PrasunBiswas nope $\endgroup$ – Nick Chapman Apr 8 '15 at 7:06
  • $\begingroup$ On Wikipedia, there is an asymptotic approximation which says $\lim_{n\rightarrow\infty} \frac{\Gamma(n+\alpha)}{\Gamma(n)n^{\alpha}} = 1$ for $\alpha \in \mathbb{R}$. That seems like it'd do the trick to show that the limit is either 1, 0 or infinity depending on the value of $a$ and $b$? $\endgroup$ – JessicaK Apr 8 '15 at 7:11
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We have $$e^{b-a}\lim_{x\rightarrow\infty}x^{b-a}\frac{\left(x+a\right)^{x+a}}{\left(x+b\right)^{x+b}}=e^{b-a}\lim_{x\rightarrow\infty}\frac{\left(x+a\right)^{a}}{x^{a}}\frac{x^{b}}{\left(x+b\right)^{b}}\frac{\left(x+a\right)^{x}}{\left(x+b\right)^{x}}=e^{b-a}\lim_{x\rightarrow\infty}\frac{\left(1+\frac{a}{x}\right)}{\left(1+\frac{b}{x}\right)^{b}}^{a}\left(\frac{x+a}{x+b}\right)^{x}=e^{b-a}\lim_{x\rightarrow\infty}\left(1+\frac{a-b}{x+b}\right)^{x}=e^{b-a}e^{a-b}=1.$$

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  • $\begingroup$ Can you elaborate on how the $x^{b-a}$ went away? $\endgroup$ – Nick Chapman Apr 8 '15 at 7:25
  • $\begingroup$ @NickChapman I added some information. $\endgroup$ – Marco Cantarini Apr 8 '15 at 7:29
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Continuing in the same spirit as Circonflexe, we could show how the limit is approached.

To the next order, we have $$\Big(1+\frac ax\Big)^{x+a}=e^a+\frac{a^2 e^a}{2 x}+O\left(\left(\frac{1}{x}\right)^2\right)$$ from which $$F(x)=x^{b-a}\frac{\Gamma(x+a+1)}{\Gamma(x+b+1)}=1+\frac{(a-b) (a+b+1)}{2 x}+O\left(\left(\frac{1}{x}\right)^2\right)$$

With regard to the approximation in the Wikipedia page, we could show, using the same way that, for large values of $n$, $$\frac{\Gamma(n+a)}{\Gamma(n)~n^a}=1+\frac{(a-1) a}{2 n}+\frac{(a-2) (a-1) a (3 a-1)}{24 n^2}+O\left(\left(\frac{1}{n}\right)^3\right)$$

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You can treat this as a rational fraction by simplifying out all the $x$, whose powers do cancel out: you get $$F(x) \sim (1+a/x)^{x+a} (1+b/x)^{-x-b}.$$ We would now simplify this by using the limited series expansion of $(1+t)^u$, but here $u$ is a function of $x$, so we must first show that this is valid. Namely, we have $$\begin{split} (1+a/x)^{x+a} &= \exp ((x+a) \log (1+a/x)) = \exp ((x+a) (a/x+O(x^{-2})) \\ &= \exp (a + O(1/x))\end{split},$$ so that unless I am mistaken, $$F(X) \sim e^{b-a} \exp(a+O(1/x)) \exp(-b+O(1/x)) \rightarrow 1$$ as $x \rightarrow +\infty$.

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Just a disclaimer that I've learned everything I know about asymptotic expansions from reading random solutions on this very site.

Since (from the wikipedia entry on the Gamma function)

$$\lim_{n\rightarrow \infty} \frac{\Gamma(n+\alpha)}{\Gamma(n)n^{\alpha}} = 1, \quad \alpha \in \mathbb{R}.$$

Write

\begin{align*} \lim_{x\rightarrow} F(x)&= \lim_{x\rightarrow \infty} x^{b-a}\frac{\Gamma(x+a+1)}{\Gamma(x+b+1)}\\ &=\lim_{n\rightarrow \infty} n^{b-a}\frac{\Gamma(n+a+1)}{\Gamma(n+b+1)}\\ &= \lim_{n\rightarrow\infty} n^{b-a}\frac{\Gamma(n)n^{a+1}}{\Gamma(n)n^{b+1}}\\ &= n^{b-a}n^{a+1}n^{-b-1}\\ &= 1. \end{align*}

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  • $\begingroup$ You should link the Wikipedia entry to your answer since I don't think this identity is well-known. $\endgroup$ – Prasun Biswas Apr 8 '15 at 7:27
  • $\begingroup$ @JessicaK I am totally with you except for where the identity $\Gamma(n+a+1)=\Gamma(n)n^a$ comes from. I'll believe you that it exists, I just can't see it clearly on the Wikipedia page. $\endgroup$ – Nick Chapman Apr 8 '15 at 7:31
  • $\begingroup$ @NickChapman Unfortunately, I don't have a reference. I am about 90% sure it can be found in Oldham and Spanier's An Atlas of Functions, and I would be surprised if it wasn't in Artin's The Gamma Function, but it is currently nearly 3 am where I am and the campus library is closed. $\endgroup$ – JessicaK Apr 8 '15 at 7:34
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    $\begingroup$ @NickChapman I didn't say they are equal, they grow asymptotically at the same rate. $\endgroup$ – JessicaK Apr 8 '15 at 7:35
  • $\begingroup$ @NickChapman I just want to follow up that the derivation of this asymptotic formula is indeed discussed in Oldham and Spanier, I didn't check Artin's book. $\endgroup$ – JessicaK Apr 10 '15 at 23:39

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