I have to check for which $x$ the series converges/diverges.

$\sum\limits_{n=1}^\infty\frac{n!}{n^n} \times (5x)^n$

I know that for $|x| < \frac{1}{5}e$ it converges and for $|x| > \frac{1}{5}e$ it diverges by using the ratio test $\frac{a_{n+1}}{a_n}$. However, this test does not tell me anything about $|x| = \frac{1}{5}e$.

How do I prove that the series diverges for $|x| = \frac{1}{5}e$ ? (Wolfram Alpha told me so)

I thought that it has something to do with $\frac{n!}{x^n}$ which would be $\frac{1}{e^x}$, but I just failed to find a proper proof.

Any help would be appreciated!

  • $\begingroup$ You don't need that ugly $\;\times\;$ there... $\endgroup$ – DonAntonio Dec 4 '13 at 12:47
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    $\begingroup$ Stirling's formula gives you a bit more than $n! \approx \left(\frac{n}{e}\right)^n$. The "more" settles it. $\endgroup$ – Daniel Fischer Dec 4 '13 at 12:52
  • $\begingroup$ @Ethan, how what you wrote helps here? $\endgroup$ – DonAntonio Dec 4 '13 at 12:53
  • $\begingroup$ @DonAntonio Using an adjusted version of the ratio test you can show it diverges. $\endgroup$ – Ethan Dec 4 '13 at 12:54
  • $\begingroup$ Could you please specify a little more, @Ethan? What "adjusted version"? What you wrote shows that $\;\frac{a_{n+1}}{a_n}\xrightarrow[n\to\infty]{}1\;$ ... $\endgroup$ – DonAntonio Dec 4 '13 at 12:57

$$x=\frac e5\implies\;\text{we have the series}\;\;\sum_{n=1}^\infty\frac{n!5^ne^n}{n^n5^n}=\sum_{n=1}\frac{n!e^n}{n^n}$$

and now you can use Stirling's Approximation

$$n!\sim\frac{n^n}{e^n}\sqrt{2\pi n}$$

so our series behaves asimptotically (for large values of $\;n\;$) as the series

$$\frac{n^n}{e^n}\sqrt{2\pi n}\frac{e^n}{n^n}=\sqrt{2\pi n}$$

and thus clearly our series diverges.

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    $\begingroup$ Thanks a lot man, you saved my day! $\endgroup$ – Christian Schnorr Dec 4 '13 at 14:48

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