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Can someone help me evaluate this limit?

$$\lim_{x\to +\infty}\frac {\zeta(1+\frac 1x)}{\Gamma(x)}$$

I never came across this kind of limit so I don't even know where to start.

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    $\begingroup$ $\zeta(s)$ has a simple pole at $s=1$ with residue $1$, so $\zeta(1+\frac{1}{x})\sim x$ as $x\to\infty$. $\endgroup$ – anon Jun 22 '15 at 17:07
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    $\begingroup$ i think the searched limit is zero $\endgroup$ – Dr. Sonnhard Graubner Jun 22 '15 at 17:32
  • $\begingroup$ As an aside, $~\displaystyle\lim_{x\to0}\frac{\zeta(1+x)}{\Gamma(x)}=1$. $\endgroup$ – Lucian Jun 22 '15 at 20:57
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Good old L'Hôpital tells us $$\lim_{x\to\infty}\frac{\zeta\left(1+\frac{1}{x}\right)}{\Gamma(x)}=\lim_{x\to\infty}\frac{\zeta^{'}\left(1+\frac{1}{x}\right)}{\Gamma^{'}(x)}=\lim_{x\to\infty}\frac{\zeta^{'}\left(1+\frac{1}{x}\right)}{\Gamma(x)\psi^{(0)}(x)},$$ where $\psi^{(n)}$is the $n$-th derivative of the digamma function.

Combining $\lim\limits_{x\to1}\zeta(x)=\lim\limits_{x\to+\infty}\Gamma(x)=\lim\limits_{x\to+\infty}\psi^{(0)}(x)=+\infty$ and $\lim\limits_{x\to1}\zeta'(x)=-\infty$ we deduce that for large enough $x$ the leftmost fraction is positive whereas the rightmost is negative, so if the limit exists it must be equal to $0$.


Alternatively, the result is straightforward rewriting $\zeta$ and $\Gamma$ by means of the Riemann functional equation, Stirling's approximation and the infinite product definition for the gamma function (due to Euler himself): $$\displaystyle\Gamma(x)=\frac{1}{x}\prod\limits_{n=1}^\infty\frac{\left(1+\frac{1}{n}\right)^x}{1+\frac{x}{n}}.$$

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  • $\begingroup$ Hopital is very powerful even dealing with special functions, thanx for your help. If you have any idea on how to do it without (just for curiosity) I'd be grateful if you'll incorporate it in the answer. $\endgroup$ – Renato Faraone Jun 22 '15 at 18:51
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    $\begingroup$ I'll see it done if I have spare time soon. And, you're welcome! $\endgroup$ – Vincenzo Oliva Jun 22 '15 at 20:27
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    $\begingroup$ @Renato I've added an alternative approach. $\endgroup$ – Vincenzo Oliva Jun 23 '15 at 6:45
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    $\begingroup$ very elegant approach, thanks :) $\endgroup$ – Renato Faraone Jun 23 '15 at 6:50
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Hint: Note that for $x>1$, we have $$\zeta \left(1+\frac{1}{x}\right) = \frac{1}{1^{1+\frac{1}{x}}}+\frac{1}{2^{1+\frac{1}{x}}}+\frac{1}{3^{1+\frac{1}{x}}}+\frac{1}{4^{1+\frac{1}{x}}}+\frac{1}{5^{1+\frac{1}{x}}}+\cdots< \\ \frac{1}{1^{1+\frac{1}{x}}}+\frac{1}{2^{1+\frac{1}{x}}}+\frac{1}{2^{1+\frac{1}{x}}}+\frac{1}{4^{1+\frac{1}{x}}}+\frac{1}{4^{1+\frac{1}{x}}}+\cdots = \\ 1+\frac{2}{2^{1+\frac{1}{x}}}+\frac{4}{4^{1+\frac{1}{x}}}+\frac{8}{8^{1+\frac{1}{x}}}+\cdots = \\ 1+\frac{1}{2^{\frac{1}{x}}}+\frac{1}{4^{\frac{1}{x}}}+\frac{1}{8^{\frac{1}{x}}}+\cdots= \\ (2^{-\frac{1}{x}})^0+(2^{-\frac{1}{x}})^1+(2^{-\frac{1}{x}})^2+(2^{-\frac{1}{x}})^3+\cdots = \\ \frac{1}{1-2^{-\frac{1}{x}}}$$

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