# Is there a standard way to compute $\lim\limits_{n\to\infty}(\frac{n!}{n^n})^{1/n}$?

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.)

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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. –  Ramey May 7 '12 at 20:00
@Ramey, Not at all :) –  Norbert May 7 '12 at 20:09

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

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$a_n \gt 0$.... –  Aryabhata May 7 '12 at 19:51
Thanks. That's a great trick, I see how to compute it now. But why does that implication follow? –  Ramey May 7 '12 at 19:54
This is a standard theorem and should be available in most textbooks, but you can also see this answer: math.stackexchange.com/questions/116183/… –  Aryabhata May 7 '12 at 19:55
@Ramey It's in Portuguese but you can have a look at this post Also pp. 28 to 31 of this pdf. –  Américo Tavares May 7 '12 at 22:53

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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Thank you, I appreciate this method. –  Ramey May 7 '12 at 20:01
@did: not bad (+1) –  Chris's sis the artist Aug 5 '12 at 13:50

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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