Evaluating $\lim_{n \to \infty}\frac{1}{n}(n!)^{\frac{1}{n}}$

Asked Dec 04 '16 at 06:15

Active Dec 04 '16 at 07:51

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I'm trying to find and prove the value of$$\lim_{n \to \infty}\frac{1}{n}(n!)^{\frac{1}{n}}$$

I was thinking that since $$\frac{1}{n}(n!)^{\frac{1}{n}} = \frac{1}{n} \left[ (1)^{\frac{1}{n}}(2)^{\frac{1}{n}}...(n)^{\frac{1}{n}} \right]$$

and we know that $$\lim_{n\to \infty}n^{\frac{1}{n}} = 1$$ and $$k^{\frac{1}{n}} \leq \ n^{\frac{1}{n}} \ \ \ \ \forall k \leq n$$

So $(n!)^{\frac{1}{n}} \leq 1 \ \forall n \ $* Then it is bounded. We also know that $\lim \frac{1}{n} =0$, therefore $$\lim_{n \to \infty}\frac{1}{n}(n!)^{\frac{1}{n}} = 0 **$$

I'm pretty sure this line of reasoning is ok. Now proving it is another thing. Any suggestions?

*I realize now that this statement is not true, but that did not get me any closer to solving the problem. I do believe that it is bounded though.

**LOL, I don't believe this to be true anymore either, a wild guess tells me that the solution may be $\frac{1}{e}$

edited Dec 04 '16 at 07:51

Matthew Conroy

11,234

asked Dec 04 '16 at 06:15

Mike

1

if you use Stirling's approximation $n! \sim \frac {n^n}{e^n}\sqrt {2 \pi n}$ you get a different result. – Ross Millikan Dec 04 '16 at 06:22
@Ross That technique is beyond the scope of my class. – Mike Dec 04 '16 at 06:52
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From Ross' comment you can deduce that the limit does not exist because the sequence has no upper bound. So you should try to show this. – miracle173 Dec 04 '16 at 07:21
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See this. – David Mitra Dec 04 '16 at 07:41
I found the solution, here is a link in case you are interested. @kobe uses the theorem I'm interested in for his second solution – Mike Dec 04 '16 at 08:58
@miracle173: the limit does exist and it is $\frac 1e$. The $n$th root takes the leading fraction to $\frac ne$, then dividing by $n$ gives $\frac 1e$. The square root term goes to $1$ because of the $n$th root. – Ross Millikan Dec 04 '16 at 14:48
I missed the $\frac{1}{n}$ factor in the expression. – miracle173 Dec 04 '16 at 16:59

Evaluating $\lim_{n \to \infty}\frac{1}{n}(n!)^{\frac{1}{n}}$

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