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

I don't know how to start. Hints are also appreciated.

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    $\begingroup$ See this. $\endgroup$ – David Mitra Feb 14 '16 at 9:32
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let $$y= \frac{(n!)^{1/n}}{n}.$$ $$\implies y=\left (\frac{n(n-1)(n-2)....3.2.1}{n.n.n....nnn}\right)^\frac{1}{n} $$ Note that we can distribute the $n$ in the denominator and give an $'n'$ to each term $$\implies \log y= \frac {1}{n}\left(\log\frac{1}{n}+\log\frac{2}{n}+...+\log\frac{n}{n}\right)$$ applying $\lim_{n\to\infty}$ on both sides, we find that R.H.S is of the form

$$ \lim_{n\to\infty} \frac{1}{n} \sum_{r=0}^{r=n}f \left(\frac{r}{n}\right)$$ which can be evaluated by integration $$=\int_0^1\log(x)dx$$ $$=x\log x-x$$ plugging in the limits (carefully here) we get $-1$.

$$ \log y=-1,\implies y=\frac{1}{e}$$

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$$ f_n=\frac{e^{(1/n)\log n!}}{n} $$ and the use Stirling's approximation $\log n!\sim n\log n -n$ to conclude $f_n\sim \frac{e^{\log n-1}}{n}\to 1/e$.

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If you know Stirling

$$n!\sim\sqrt{2\pi n}\left({n\over e}\right)^n$$

You get

$$u_n\sim \left(\sqrt{2\pi n}\right)^{1\over n}\cdot {1\over e}$$

Now at $+\infty$ one has $\left(\sqrt{2\pi}\right)^{1\over n}\to 1$ and $n^{1\over n}\to 1$ (look at the logarithm) and so

$$u_n\to {1\over e}$$

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Let $f_n=\frac{n!}{n^n}$ so that $f_{n+1}= \frac{(n+1)!}{(n+!)^n+1}$

So $\frac{f_{n+1}}{f_n}=(1+\frac{1}{n})^{-n}$

Therefore

$\lim_{n\to\infty}(1+\frac{1}{n})^{-n}=\frac{1}{e}>0 $

We know that if $<f_n>$ is a sequence such that $f_n$ is greater than 0 for all n, and $\lim_{n\to\infty}\frac{f_{n+1}}{f_n}=l, l>0$. Then $\lim_{n\to\infty}{f_n}^{\frac{1}{n}}=l$.

Using the above theorem we have $\lim_{n\to\infty} \frac{(n!)^{\frac{1}{n}}}{n}=\frac{1}{e}$

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