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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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    $a_n \gt 0$....2012-05-07
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    Thanks. That's a great trick, I see how to compute it now. But why does that implication follow?2012-05-07
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    This is a standard theorem and should be available in most textbooks, but you can also see this answer: http://math.stackexchange.com/questions/116183/show-that-this-limit-is-equal-to-liminf-a-n1-n-for-positive-terms/116198#1161982012-05-07
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    @Ramey It's in Portuguese but you can have a look at [this post](http://problemasteoremas.wordpress.com/2008/10/16/limite-da-raiz-de-indice-n-do-termo-geral-de-uma-sucessao/) Also pp. 28 to 31 of [this](http://problemasteoremas.files.wordpress.com/2007/11/cadernoproblemasteoremas6junho2009.pdf) pdf.2012-05-07
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    http://math.stackexchange.com/questions/116183/show-that-this-limit-is-equal-to-liminf-a-n1-n-for-positive-terms/153894#1538942012-06-22
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    @AmihaiZivan: 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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    I didn't know the second until now, thanks.2012-05-07
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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