Given the series $\sum^{\infty}_{n=1} \frac{n!z^n}{n^n}$ find the radius of convergence.

Well, I know that if the following Lemma holds:


Given the series $\sum^{\infty}_{n=1} c_nz^n$ where $c_n,z \in \mathbb C$. If $\beta := lim\frac{|c_n|}{|c_{n+1}|}$ exists, then it is also agrees with the radius of convergence $\rho := \frac{1}{limsup|c_n|^{1/n}}$. That is, $\beta = \rho$

Attempt at the Solution

First I apply the lemma above:

$\frac{\frac{n!}{n^n}}{\frac{(n+1)!}{n^{n+1}}} = \frac{n}{n+1} = 1 -\frac{1}{n+1} \rightarrow 1$

If I now consider

$\limsup(\frac{n!}{n^n})^{1/n} = \limsup\frac{(n!)^{1/n}}{n} \rightarrow +\infty$

Since $\limsup(n!)^{1/n} \rightarrow +\infty$ faster than $\limsup n \rightarrow +\infty$

On the authority of the textbook I'm reading, the radius of convergence should be $e$ I'm not quite sure how one comes to that conclusion. If anyone could point me into the right direction that would be great. Thanks.

  • 2
    $\begingroup$ The denominator is $\dfrac{(n+1)!}{(n+1)^{n+1}}$, not $\dfrac{(n+1)!}{n^{n+1}}$. $\endgroup$ – Daniel Fischer Oct 12 '15 at 20:45

Use Stirling's approximation

$$n! \sim n^n e^{-n} \sqrt{2 \pi n} \quad (n \to \infty)$$

You should be able to conclude that the radius of convergence is $e$.

  • $\begingroup$ The Stirling's approximation works great, but is there another method to find e? $\endgroup$ – docjock Oct 12 '15 at 23:35

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