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

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

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

Community
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user1003365
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2 Answers2

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

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

so the limit is

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

Ron Gordon
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  • 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. – user1003365 Jun 20 '15 at 19:51
  • @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. – Ron Gordon Jun 20 '15 at 19:53
  • I'm still not convinced... You can have functions that are continuous and still the Taylor series doesn't converge to the value. http://math.stackexchange.com/questions/721364/why-dont-taylor-series-represent-the-entire-function/721379#721379 – user1003365 Jun 20 '15 at 20:07
  • @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. – Ron Gordon Jun 20 '15 at 20:11
  • 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).

     – user1003365 Jun 20 '15 at 20:38
  • @RonGordon +1 on your approach! Another point to make here is that you are not using the Taylor series for $\Gamma$. Rather, for $x$ real-valued, you are effectively using the extended mean-value theorem (i.e., $\Gamma(x+\epsilon)=\Gamma(x)+\Gamma'(x)\epsilon+\frac12 \Gamma''(\xi)\epsilon^2$, for some $x<\xi <x+\epsilon$). So, convergence of a series is not even an issue. – Mark Viola Jun 20 '15 at 20:41
  • Hi MV, do you mean this one? https://en.wikipedia.org/wiki/Mean_value_theorem_(divided_differences) I don't see how $\Gamma'$ can make its way there when you consider $n=2$. – user1003365 Jun 20 '15 at 20:55
  • OK, just to clarify, @RonGordon 's answer was helpful and seems to correct to me if we take into account that the function is analytic. I can't upvote as I don't have enough reputation yet :( – user1003365 Jun 20 '15 at 21:18
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{\lim_{x \to \infty}\,\,{\Gamma\pars{x + 1} \over \Gamma\pars{x + 1 + 1/x^{2}}}} = \lim_{x \to \infty}\,\,{x! \over \pars{x + 1/x^{2}}!} \\[5mm] = &\ \lim_{x \to \infty}\,\,{\root{2\pi}x^{x + 1/2}\,\,\expo{-x} \over \root{2\pi}\pars{x + 1/x^{2}}^{x\ +\ 1/x^{2}\ + 1/2} \,\,\,\,\expo{-x - 1/x^{2}}\,\,} \\[5mm] = &\ \lim_{x \to \infty}\,\,{x^{x + 1/2} \over x^{x\ +\ 1/x^{2}\ + 1/2}\,\,\, \bracks{\pars{1 + 1/x^{3}}^{x^{3}}}^{1/x^{2}\ +\ 1/x^{5}\ + 1/\pars{2x^{3}}} \,\,\,\,}\,\expo{1/x^{2}} \\[5mm] = & \lim_{x \to \infty}x^{-1/x^{2}} = \exp\pars{-\lim_{x \to \infty}{\ln\pars{x} \over x^{2}}} = \exp\pars{-\lim_{x \to \infty}{1/x \over 2x}} \\[5mm] = & \bbx{1} \\ & \end{align}

Felix Marin
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  • Thank you so much for providing an answer. I had figured something myself but your answer is far cleaner. If we agree on something I can mention you if I ever get to publish the research where I'm using it. The only thing I don't get... Your reasoning seems to assume (or at least imply?) that $(1 + 1/x^3)^{x^3}^(1/x^2 + 1/x^5 + 1/(2x^{3}))$ goes to 1 (as $e^{1/x^{2}}$ goes to 1 and hence things cancel giving the equality). However I can't see that: as $1 + 1/x^{3}$ goes to 1, but $x^3$ goes to infinity, and $1^\inf$ is indeterminate. What am I missing? – user1003365 Apr 21 '21 at 10:42