find the limit of the sequence: (n!) ^ (1/n) as n -> infinity

i don't even know where to start on this one. i think if i get a nod in the right direction i'd be able to get it, so can anyone here give me that nod?

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- Nov 30th 2007, 04:11 PMbartendorsparkyfind the limit of the sequence
find the limit of the sequence: (n!) ^ (1/n) as n -> infinity

i don't even know where to start on this one. i think if i get a nod in the right direction i'd be able to get it, so can anyone here give me that nod? - Nov 30th 2007, 06:31 PMTKHunny
It could help to consider $\displaystyle \frac{1}{n}ln(n!)\;=\;\frac{1}{n}(ln(n)+ln(n-1)+...+ln(2))$

- Nov 30th 2007, 06:46 PMPaulRS
$\displaystyle \sqrt[n]{n!}\approx{\frac{n}{e}}$ as $\displaystyle n\rightarrow{+\infty}$

Simply: $\displaystyle e^n=1+n+\frac{1}{2}n^2+...+\frac{n^n}{n!}+...$

Thus: $\displaystyle e^n\geq{\frac{n^n}{n!}}$ and $\displaystyle n!\geq{\frac{n^n}{e^n}}$ (this is enough for what you want to know, but do go on)

And: $\displaystyle \int_1^{n+1}{\ln(x)dx}\geq{\sum_{k=1}^{n}{\ln(i)}} =\ln(n)!$

Then $\displaystyle (n+1)\ln(n+1)-n\geq{\ln(n)!}$ and $\displaystyle \frac{(n+1)^{(n+1)}}{e^n}\geq{n!}$

So we have $\displaystyle \frac{(n+1)^{n+1}}{e^n}\geq{n!}\geq{\frac{n^n}{e^n }}$

Therefore: $\displaystyle \frac{(n+1)^{1+\frac{1}{n}}}{e}\geq{\sqrt[n]{n!}}\geq{\frac{n}{e}}$

Now if you divide by $\displaystyle \frac{n}{e}$ and take the limit you get that, by the squeeze theorem $\displaystyle \sqrt[n]{n!}\approx{\frac{n}{e}}$ - Dec 1st 2007, 02:27 PMThePerfectHacker
Use the following trick.

Let $\displaystyle a_n$ be a seqeuence of non-zero real numbers. If $\displaystyle \left| \frac{a_{n+1}}{a_n}\right|$ converges then $\displaystyle \left| a_n \right|^{1/n}$ converges to the same thing.