Wednesday, 13 May 2015

analysis - Evaluating limntoinftyfrac1n(n!)frac1n

I'm trying to find and prove the value oflim




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}

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