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

Question

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.)

See also: Finding the limit of $\frac {n}{\sqrt[n]{n!}}$ and other posts linked there. — Martin Sleziak, May 31 '20 at 21:53

score 7 · Answer 1 · answered May 07 '12 at 19:50

7

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

answered May 07 '12 at 19:50

Aryabhata

82,206

1

$a_n \gt 0$.... – Aryabhata May 07 '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 07 '12 at 19:54
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#116198 – Aryabhata May 07 '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 07 '12 at 22:53
http://math.stackexchange.com/questions/116183/show-that-this-limit-is-equal-to-liminf-a-n1-n-for-positive-terms/153894#153894 – Amihai Zivan Jun 22 '12 at 08:16
@AmihaiZivan: Thanks. – Aryabhata Jun 22 '12 at 19:17

score 6 · Accepted Answer · answered May 07 '12 at 19:51

6

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.

answered May 07 '12 at 19:51

Norbert

56,803

I didn't know the second until now, thanks. – Ramey May 07 '12 at 20:00
@Ramey, Not at all :) – Norbert May 07 '12 at 20:09

Did · Answer 3 · 2012-05-08T06:10:11.053

4

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, $\lim\limits_{n\to\infty}x_n=1/\mathrm e$.

edited May 08 '12 at 06:10

answered May 07 '12 at 19:55

Did

279,727

@did: not bad (+1) – user 1591719 Aug 05 '12 at 13:50

score 2 · Answer 4 · answered May 07 '12 at 20:03

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.

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

4 Answers4

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