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$$\lim_{k\rightarrow\infty}(k!)^{\frac{1}{k}}$$ I want to determine the limit as formal as possible, that is, to prove, for example $$\lim_{k\rightarrow\infty}\frac{\ln(k!)}{k}$$ I had the intention of using the L'Hopital rule and I do not know that it does exist, despite only representing a number. But it is not very justified.

Can you use sanduche theorem? Can I limit?

Asaf Karagila
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Santiago Seeker
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3 Answers3

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Hint: for a sequence of positive numbers $a_n$, if $\lim_{n\rightarrow\infty}(\frac{a_{n+1}}{a_n})$ exists and equals $L$ ($L$ can be a real number of infinity, it doesn't matter) then $\lim_{n\rightarrow\infty}(\sqrt[n]{a_n})=L$.

Mark
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HINT:

Note that

$$\begin{align} \frac{\log(k!)}k&=\frac1k\sum_{n=1}^k \log(n)\\\\ &=\log(k)+\frac1k\sum_{n=1}^k \log(n/k)\tag1 \end{align}$$

The sum on the right-hand side of $(1)$ is the Riemann sum for the improper integral $\lim_{\varepsilon\to0}\int_\varepsilon^1 \log(x)\,dx$.

Can you finish this?

Mark Viola
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  • Oh, wow! that great ... is the criterion of the integral and with that I show that it is divergent :) thank you very much !! – Santiago Seeker Aug 05 '18 at 18:57
  • You're welcome. My pleasure. – Mark Viola Aug 05 '18 at 19:11
  • Avoiding the problem of why those Riemann sums converge to the improper integral, we could estimate

    $$\sum_{j=1}^{k}\ln j \ge (k/2)\ln (k/2),$$

    from which the result follows.

     – zhw. Aug 05 '18 at 22:29
  • No concavity needed, just need  $\ln j$ increasing. And I'm doing fine thank you, and I hope the same is true your way. – zhw. Aug 05 '18 at 22:40
  • I see. You're using $n!\ge (n/2)^{n/2}$ and $\log(x)$ is increasing. I missed that upon first glance. – Mark Viola Aug 05 '18 at 22:54
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As an alternative by Stolz-Cesaro

$$\lim_{k\rightarrow\infty}\frac{\ln(k!)}{k}=\lim_{k\rightarrow\infty}\frac{\ln((k+1)!)-\ln(k!)}{k+1-k}=\lim_{k\rightarrow\infty} \log (k+1) =\infty$$

user
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