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I think I've discovered a new definition for the Euler-Mascheroni Constant (Gamma)

I can't find it online anywhere, has anyone seen it before?

$$\lim_{x \to 0} (-\ln ( \sqrt[x]{x!} )) = \gamma$$

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    $\begingroup$ This is essentially the fact that $ -\gamma = \Gamma'(1)$. $\endgroup$
    – lhf
    Oct 1, 2014 at 2:09
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    $\begingroup$ Your limit is the definition of $-\psi_0(1)$, which is $(48)$ on MathWorld. $\endgroup$
    – anon
    Oct 1, 2014 at 2:09
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    $\begingroup$ Here is a post proving @lhf's approach. $\endgroup$
    – Lucian
    Oct 1, 2014 at 3:04

3 Answers 3

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Write $$ \ln ( \sqrt[x]{x!} ) = \ln ({x!}^{1/x} ) = \frac{\ln ({x!})}{x} = \frac{\ln (\Gamma(x+1))}{x} = \frac{f(x)-f(0)}{x-0} $$ for $$ f(x)= \ln (\Gamma(x+1)) $$ Then $$ \lim_{x \to 0} \ln ( \sqrt[x]{x!}) = f'(0) $$ Now $$ f'(x)= \frac{\Gamma'(x+1)}{\Gamma(x+1)} $$ and so $$ f'(0) = \frac{\Gamma'(1)}{\Gamma(1)} = \Gamma'(1) = -\gamma $$ The last equality is the only difficult part, but it is a standard fact.

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The Limit

The recurrence $\Gamma(x+1)=x\Gamma(x)$ gives us that $$ \begin{align} \left(\frac{\Gamma(x+n+1)}{\Gamma(n+1)}\right)^{1/x} &=\Gamma(x+1)^{1/x}\left(\frac{(x+1)(x+2)\cdots(x+n)}{n!}\right)^{1/x}\\ &=\Gamma(x+1)^{1/x}\left(1+\frac x1\right)^{1/x}\left(1+\frac x2\right)^{1/x}\cdots\left(1+\frac xn\right)^{1/x}\tag{1} \end{align} $$ As shown below, the log-convexity of $\Gamma(x)$ guarantees that $$ (n+x)^x<\frac{\Gamma(x+n+1)}{\Gamma(n+1)}<(n+1)^x\tag{2} $$ Combining $(1)$ and $(2)$ gives $$ 1\lt\lim_{x\to0}\Gamma(x+1)^{1/x}e^{H_n-\log(n)}\lt 1+\frac1n\tag{3} $$ Because $n$ is arbitrary, the Squeeze Theorem yields $$ \lim_{x\to0}\Gamma(x+1)^{1/x}=e^{-\gamma}\tag{4} $$ which is a restatement of the question.


Log-Convexity of $\,\boldsymbol{\Gamma(x)}$

Since $\Gamma(x)$ is log-convex, we have $$ \begin{align} \Gamma(x+n+1) &\le\Gamma(n+1)^{1-x}\Gamma(n+2)^x\\ &=\Gamma(n+1)^{1-x}[(n+1)\Gamma(n+1)]^x\\ &=\Gamma(n+1)(n+1)^x\tag{5} \end{align} $$ and $$ \begin{align} \Gamma(n+1) &\le\Gamma(x+n)^x\Gamma(x+n+1)^{1-x}\\ &=\left[\frac{\Gamma(x+n+1)}{n+x}\right]^x\Gamma(x+n+1)^{1-x}\\ &=\Gamma(x+n+1)(n+x)^{-x}\tag{6} \end{align} $$ Combining $(5)$ and $(6)$ yields $(2)$.

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More in general we have

$${a}^{-1}b\,\lim _{x\rightarrow 0}(-\ln \left( \left( \left( a\,x \right) ! \right) ^{{\frac {1}{b\,x}}} \right)) =\gamma $$

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    $\begingroup$ while this is true, it would be nice to give some explanation. $\endgroup$
    – robjohn
    Oct 1, 2014 at 5:25

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