Can someone please help me with the calculation of this limit?

$\lim_{x \to \infty} \Gamma(x+1)/\Gamma(x+1+1/x^2)$

I tried wolframalpha and seems to be 1, yet there are no "detailed steps" as to how it reaches the conclusion (I understand there wouldn't be even if I pay, as in other cases it shows me the first ones).

Thanks in advance, Sergio

  • $\begingroup$ Have you tried Stirling's approximation? $\endgroup$ – Chappers Jun 20 '15 at 19:20
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    $\begingroup$ Heuristically, when $x$ becomes very large, does the $1/x^2$ term matter? $\endgroup$ – Nigel Overmars Jun 20 '15 at 19:21

$$\Gamma(z+\epsilon) =\Gamma(z) + \epsilon \Gamma'(z) + O(\epsilon^2) $$

$$\Gamma'(z) = \Gamma(z) \psi(z) $$

so the limit is

$$\lim_{x \to \infty} \frac1{1+ \psi(x+1)/x^2} = \lim_{x \to \infty} \frac1{1+ (\log{x}-\gamma)/x^2} = 1$$

  • $\begingroup$ Thanks for the super quick answer. You seem to be using the Taylor series for $\Gamma$. Is it known to be an entire function? I can find references to the fact that the inverse Gamma is, but not the function itself. Thanks again. $\endgroup$ – user1003365 Jun 20 '15 at 19:51
  • $\begingroup$ @user1003365: doesn't have to be entire - just sufficiently far from a pole. Given that the poles are all negative integers and zero, and you are taking the limit for large $x$, I think that is a safe assumption. $\endgroup$ – Ron Gordon Jun 20 '15 at 19:53
  • $\begingroup$ I'm still not convinced... You can have functions that are continuous and still the Taylor series doesn't converge to the value. math.stackexchange.com/questions/721364/… $\endgroup$ – user1003365 Jun 20 '15 at 20:07
  • $\begingroup$ @user1003365: You are completely ignoring the specific function here. Yes, it is true that a Taylor series does not necessarily represent the entire function in the complex plane unless it is entire. But...it does represent it away from a pole. The example you show is a function with an essential singularity at zero, so of course it doesn't have a Taylor series about that point! But here we have the gamma function for large positive $x$, a region in which the Gamma function has no singularities whatsoever. The Taylor expansion is valid for what we have. $\endgroup$ – Ron Gordon Jun 20 '15 at 20:11
  • $\begingroup$ I'm seeing that $\Gamma$ has an analytic continuation to the complex plane except for negative integers. Since we are looking at values $x>1$, it seems it is analytic for the values we are interested in... (As a separate example, sorry, my example wasn't very illustrative). $\endgroup$ – user1003365 Jun 20 '15 at 20:38

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