Show that the following sequence converges for $ 0 < a < e $ and diverges for $ a \ge e$

Question

I have this question which I'm having trouble solving, can I use some help? :)

Show that the following sequence converges for $ 0 < a < e $ and diverges for $ a \ge e$:

$ \sum_{n=1}^{\infty} \frac{a^nn!}{n^n} $

Thanks a lot!

Try to find the radius of convergence as $R = \lim\limits_{n\to\infty} \left| \dfrac{c_n}{c_{n+1}} \right|,$ where $c_n=\dfrac{n!}{n^n}.$ — M. Strochyk, May 25 '15 at 22:09
Or even better : https://math.stackexchange.com/questions/177538/finding-the-convergence-interval-of-sum-limits-n-0-infty-fracnxnnn — Arnaud D., Nov 29 '19 at 02:29

score 1 · Answer 1 · answered May 25 '15 at 22:18

1

Hint 1: Use the Ratio Test.

Note that $$\left|\frac{a^{n+1}(n+1)!}{(n+1)^{n+1}}\cdot \frac{n^n}{a^n\cdot n!}\right| = a\cdot\frac{n+1}{n}\left(\frac{n}{n+1}\right)^{n+1}.$$

Hint 2: Show that $\lim_{n\rightarrow \infty} \left(\frac{n}{n+1}\right)^{n+1} = 1/e.$

answered May 25 '15 at 22:18

user 1987

What if the limit is $1$? – David Mitra May 25 '15 at 22:23
By Stirling's theorem, $n! \approx c\sqrt{n}(n/e)^n$, so $e^n n!/n^n \approx c \sqrt{n}$ so the sum diverges. This can be proved by reasonably elementary means using the properties of log and exp. – marty cohen May 25 '15 at 22:33

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