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I am looking for a proofs of the following limits:

$$ \lim_{x \to \infty} \Gamma \left(1+\frac{1}{x} \right)^x = e^{-\gamma}. $$ I find this limit interesting as it relates the gamma function $\Gamma$ with the other gamma $\gamma$ which is the Euler-Mascheroni constant.

The second limit whose proof I am interested in is $$ \lim_{x \to 0} x \Gamma \left(1+\frac{1}{x} \right)^x = e^{-1}. $$

Nilotpal Sinha
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  • Do you know the product representation of $\Gamma$? – Daniel Fischer Aug 01 '13 at 12:36
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    The first one is just an exponentiated form of the fact that $\Gamma'(1) = -\gamma$, which is well known (just write $x = 1/h$ with $h \rightarrow 0$ and use the limit definition of the derivative for $\log \Gamma(x)$). – KCd Aug 01 '13 at 12:38
  • my edit @user60930 – what'sup Aug 01 '13 at 12:39
  • The second one (which is meant to be $x \rightarrow 0^+$, I presume) is a quick consequence of the logarithmic form of Stirling's formula. Write down Stirling's asymptotic formula for $\Gamma(t+1)$ as $t \rightarrow \infty$, take the logarithm of both sides, divide by $t$, and then write $t$ as $1/x$. – KCd Aug 01 '13 at 12:43
  • Could you tell us how you came to be looking for proofs of these results? Since they are both exponentiated versions of classical properties of the $\Gamma$-function, they sound like problems from a book or something, rather than conjectures you'd have come up with in your own calculations. – KCd Aug 01 '13 at 12:44
  • @what'sup Just to make sure you know about this - if many of your edits will be rejected you might be blocked from suggesting edits for some time. See meta. – Martin Sleziak Aug 01 '13 at 12:46
  • @MartinSleziak thank you very much i didn't know i just like $$\Large{LaTeX}$$ – what'sup Aug 01 '13 at 12:58
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    @what'sup Several suggestions: 1) do not use \Large 2) do not put all formulas between double dollars, some of them look much better in the text 3) if you change, say $x$ to $\mathrm{x}$, do it everywhere 4) avoid minor edits and try to adress all issues of the post. – Start wearing purple Aug 01 '13 at 13:10
  • Wolfram Alpha will evaluate the first of these limits for you, and give you step by step instructions http://www.wolframalpha.com/input/?i=limit+gamma%28%281%2B1%2Fx%29%29^x+x-%3EInfinity It can't sort out the second one within its default calculation time allowance though http://www.wolframalpha.com/input/?i=limit+x*gamma%28%281%2B1%2Fx%29%29^x+x-%3E0 – WalkingRandomly Aug 01 '13 at 13:33
  • Gamma and gamma are just symbols used to represent different things in mathematics...btw – zerosofthezeta Aug 09 '13 at 05:10

3 Answers3

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As shown in this answer, $\Gamma'(1)=-\gamma$. Thus, $\Gamma\left(1+\frac1x\right)=1-\frac\gamma{x}+O\left(\frac1{x^2}\right)$ and therefore, $$ x\log\left(\Gamma\left(1+\frac1x\right)\right)=-\gamma+O\left(\frac1x\right) $$ and $$ \lim_{x\to\infty}\Gamma\left(1+\frac1x\right)^{\large x}=e^{-\gamma} $$

The second question is essentially the same as $$ \lim_{n\to\infty}\frac1n(n!)^{1/n}=\frac1e $$ mentioned in this answer if we set $x=\frac1n$, since $n!=\Gamma(1+n)$.

By Stirling's Approximation, $$ n!\sim\sqrt{2\pi n}\,n^ne^{-n} $$ therefore, $$ \begin{align} \lim_{n\to\infty}\frac1n(n!)^{1/n} &=\lim_{n\to\infty}\frac1n\frac ne\lim_{n\to\infty}\sqrt{2\pi n}^{1/n}\\ &=\frac1e \end{align} $$

Community
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robjohn
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  • First limit: take the logarithm and use that $\gamma=-\psi(1)=-\Gamma'(1)$.

  • Second limit: take the logarithm and use Stirling's approximation $\ln\Gamma(1+z)=z(\ln z-1)+O(1)$ as $z\rightarrow+\infty$.

Start wearing purple
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An excellent discussion of this topic can be found in the book The Gamma Function by James Bonnar. In regards to the first question, see Chapter 8 which covers the Weierstrass product form of the Gamma function. In regards to the second question, see Chapter 15 on Stirling's formula. Both of these results are derived in the book.

james
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