How to find $\lim _{ n\to \infty } \frac { ({ n!) }^{ 1\over n } }{ n }$? [duplicate]

How to find $\lim _{ n\to \infty } \frac { ({ n!) }^{ 1\over n } }{ n }$ ? I tried taking using logarithm to bring the expression to sum form and then tried L Hospital's Rule.But its not working.Please help!

This is what wolfram alpha is showing,but its not providing the steps!

BTW if someone can tell me a method without using integration, I'd love to know!

marked as duplicate by draks ..., Community♦Jun 16 '15 at 19:12

Note

\begin{align}\frac{(n!)^{1/n}}{n} &= \left[\left(1 - \frac{0}{n}\right)\left(1 - \frac{1}{n}\right)\left(1 - \frac{2}{n}\right)\cdots \left(1 - \frac{n-1}{n}\right)\right]^{1/n}\\ &= \exp\left\{\frac{1}{n}\sum_{k = 0}^{n-1} \log\left(1 - \frac{k}{n}\right)\right\} \end{align}

and the last expression converges to

$$\exp\left\{\int_0^1\log(1 - x)\, dx\right\} = \exp(-1) = \frac{1}{e}.$$

Alternative: If you want to avoid integration, consider the fact that if $\{a_n\}$ is a sequence of positive real numbers such that $\lim\limits_{n\to \infty} \frac{a_{n+1}}{a_n} = L$, then $\lim\limits_{n\to \infty} a_n^{1/n} = L$.

Now $\frac{(n!)^{1/n}}{n} = a_n^{1/n}$, where $a_n = \frac{n!}{n^n}$. So

$$\frac{a_{n+1}}{a_n} = \frac{(n+1)!}{(n+1)^{n+1}}\cdot \frac{n^n}{n!} = \frac{n+1}{n+1}\cdot\frac{n^n}{(n+1)^n} = \left(\frac{n}{n+1}\right)^n = \left(\frac{1}{1 + \frac{1}{n}}\right)^n = \frac{1}{\left(1 + \frac{1}{n}\right)^n}.$$

Since $\lim\limits_{n\to \infty} (1 + \frac{1}{n})^n = e$, then $$\lim_{n\to \infty} \frac{a_{n+1}}{a_n} = \frac{1}{e}.$$

Therefore $$\lim_{n\to \infty} \frac{(n!)^{1/n}}{n} = \frac{1}{e}.$$

• Can you please explain why it converges to that integral ? – user220382 Jun 16 '15 at 14:29
• It's a Riemann sum for that integral – Simon S Jun 16 '15 at 14:30
• @SimonS hmmm,but I dont know Riemann sum.Can you send me a link or something? (Actually I'm a beginner in calculus..sooo...) – user220382 Jun 16 '15 at 14:31
• Have you studied integrals? The definition of a definite integral is as a Riemann sum. See, e.g., khanacademy.org/math/integral-calculus/… – Simon S Jun 16 '15 at 14:32
• Yes I studied basic integration.But I dont know about Reimann.Ok i'm checking. Thanks btw :-) @SimonS – user220382 Jun 16 '15 at 14:34

Use Stolz Cezaro:

$$\ln \lim _{ n\to \infty } \frac { ({ n!) }^{ 1/n } }{ n } =\lim _{ n\to \infty } \ln \left( \frac { ({ n!) } }{ n^n } \right)^\frac{1}{n} =\lim _{ n\to \infty } \frac{\ln(n!)- n \ln(n)}{n}$$

Now by SC we get $$\ln \lim _{ n\to \infty } \frac { ({ n!) }^{ 1/n } }{ n } =\lim _{ n\to \infty } \ln((n+1)!)- (n+1) \ln(n+1)-\ln(n!)+n\ln(n) \\= \lim _{ n\to \infty } \ln\frac{(n+1)!}{n!}- (n+1) \ln(n+1)+n\ln(n) \\=\lim _{ n\to \infty } -n\ln(n+1)+n\ln(n)=\ln \frac{1}{(1+\frac{1}{n})^n}=\ln \frac{1}{e}$$

• ...shouldn't it be $-1$ then? – draks ... Jun 16 '15 at 14:43
• @draks... Thank you fixed, forgot to write the ln. – N. S. Jun 16 '15 at 16:50

Use Stirling $n!\sim n^ne^{-n}\sqrt{2\pi n}$ to see that $$\lim_{n \to \infty} \left(\frac{n!}{n^n}\right)^{1/n} =\lim_{n \to \infty} \left(\frac{n^n e^{-n}\sqrt{2\pi n}}{n^n}\right)^{1/n} =\lim_{n \to \infty} \frac1e \left({\sqrt{2\pi n}}\right)^{1/n}=\frac1e$$

• This answer looks very similar. – robjohn Jun 16 '15 at 17:33
• @robjohn hmm... true... seems I like Stirling. A problem? – draks ... Jun 16 '15 at 18:01
• Why answer the same question twice with essentially the same answer? Why not just mark this question as a duplicate? – robjohn Jun 16 '15 at 18:21
• @robjohn I forgot my previous answer. Should I delete this one? – draks ... Jun 16 '15 at 18:47
• That is up to you. However, since this question has been asked at least 6 times before, someone should vote to close this question as a duplicate. I may do so, but if I vote to close, it will be closed unilaterally. – robjohn Jun 16 '15 at 18:55