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What is the value of $$\lim_{k\to\infty}(k!)^{\frac{1}{k}}?$$

One of my students concluded the limit was infinity – which I tend to agree with, but was unable to show that was the limit. We knew $k!$ was tough to beat, but $k^k$ does – so this situation was unclear.

Martin Sleziak
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4 Answers4

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The limit is infinity. Just note that $k! \ge (k/2)^{k/2}$ (except for very small $k$).

quid
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If we use stirling's approximation: $$n!\sim \left(\frac{n}{e}\right)^n\sqrt{2\pi n}$$ we can conclude that the limit is infinity.

Elaqqad
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We have $k! < k^k$. Further, we have $$e^k = \sum_{l=0}^{\infty} \dfrac{k^l}{l!} \implies e^k > \dfrac{k^k}{k!} \implies k! > \left(\dfrac{k}e\right)^k$$ Hence, we have $$\dfrac{k}e < (k!)^{1/k} < k$$ Now conclude.

Adhvaitha
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Just consider that $$(k!)^\frac{1}{k}=e^{ln(k!)^\frac{1}{k}}=e^\frac{ln(k!)}{k}$$

ie we have $$\lim_{k \to \infty} k!(^\frac{1}{k})= e^{\lim_{k \to \infty}\frac{ln(k!)}{k}}$$

considering just the exponent and using L'Hospital rule you see $$\lim_{k \to \infty} \frac{\ln(1 \bullet 2 \bullet … \bullet k)}{k}=\frac{\ln(2)+\ln(3)+…+\ln(k)}{k}$$

Now bound the top by an integral of $\ln(k)$ i.e., $$ln(2)+ln(3)+…+ln(k) \ge \int_{k=1}^{\infty}\ln(x)dx = k \ln k-k $$ from 2 to $\infty$ and evaluate i.e., , the k on the bottom will cancel. $\frac{k(\ln k-1)}{k}$=$\ln k-1$ which clearly $\rightarrow \infty$ as $k \to \infty$and you will have the limit to be $\infty$, thus your original limit is $$\lim_{k \to \infty}e^k=\infty$$

Quality
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    The math in this answer looks very dubious. And please don't write $e^\infty$, that's just asking for trouble. $\infty$ is not a number that can be used in algebraic expressions. I think the approach is workable, but the presentation is not very good. – Mario Carneiro Apr 10 '15 at 20:38
  • Well, if it is dubious please point out why. And are you sure it is not valid to write that..? – Quality Apr 10 '15 at 20:42
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    Your bound just now goes the wrong way - you proved that the limit is less than infinity, which is not so useful. – Mario Carneiro Apr 10 '15 at 20:43
  • $e^x$ is a function $\Bbb R\to\Bbb R$ (or $\Bbb C\to\Bbb C$ if you prefer), and $\infty\notin\Bbb R$, so $e^\infty$ is an undefined expression. To prove that the limit is $\infty$, you need to show that it is larger than any $M$, and just note that if $x$ exceeds $\log M$ then $e^x$ exceeds $M$, so if you can show that $\log{n!}/n$ is eventually larger than any constant (in particular $\log M$) then you are done. – Mario Carneiro Apr 10 '15 at 20:49
  • Okay, if any other members with experience tell me it is all incorrect than I will go ahead and delete it. Thanks – Quality Apr 10 '15 at 20:51
  • However I think the writing of e to the infinity is of less importance – Quality Apr 10 '15 at 20:53
  • It's not incorrect, in the sense that you have the right idea, but there are some mistakes in the proof. As for the integral, you should use a lower bound instead so that $\sum_{i=2}^k\log k\ge\int_1^k\log x,dx=k\log k-k+1$. – Mario Carneiro Apr 10 '15 at 20:56