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I'm computing the radii of convergence for some complex power series. For one I need to compute $\lim_{n\to\infty}\left(\frac{n!}{n^n}\right)^{1/n}.$

I know the answer is $\frac{1}{e}$, so the radius is $e$. But how could you compute this by hand? I tried taking the logarithms and raising $e$ by this logarithm, but it didn't lead me to the correct limit. (This is just practice, not homework.)

4 Answers 4

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Stirling's formula for $n!$ works wonders here, but I guess that may not qualify as "computing by hand"?

Just for the sake of completeness, Stirling's formula states that

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

or:

$\displaystyle\lim_{n\to\infty} \frac{n!}{\sqrt{2\pi n}\left(\frac{n}{e}\right)^n} = 1$

Substituting the RHS from the first equation and taking the limit as $n \to \infty$ pretty much yields the solution instantly.

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$\frac{a_{n+1}}{a_n} \to L \implies (a_n)^{1/n} \to L$

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    @Am$i$ha$i$Zivan: Thanks.2012-06-22
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There are two formulas to compute radius of convergence of the series $\sum\limits_{n=1}^\infty{c_n}z^n$ $ \frac{1}{R}=\lim\limits_{n\to\infty}|c_n|^{1/n}=\lim\limits_{n\to\infty}\left|\frac{c_{n+1}}{c_n}\right|. $ Use the second one.

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    @Ramey, Not at all :)2012-05-07
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One can indeed consider logarithms (among other methods). To see this, call the $n$th term $x_n$, then $ \log(x_n)=n^{-1}\log(n!)-\log(n)=n^{-1}\sum_{k=1}^n\log(k/n). $ This is a Riemann sum, hence, when $n\to\infty$, $ \log(x_n)\to\int_0^1\log(x)\mathrm dx=\left[x\log(x)-x\right]_0^1=-1, $ that is, $\lim\limits_{n\to\infty}x_n=1/\mathrm e$.

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    @did: not bad (+1)2012-08-05