# Show that the series converges for all $0 < a < 3$

The series is: $\sum_\limits{n=1}^\infty \dfrac {n!}{n^n}a^n$

This is what I have so far:

Using the ratio test, I have $$\lim\limits_{n \to \infty} \left|\frac {(n+1)!}{(n+1)^{n+1}}\frac {n^n}{n!}\frac {a^{n+1}}{a^n}\right|=\lim\limits_{n \to \infty} \left\vert a\left(\frac{1}{1+\frac{1}{n}}\right)^n\right\vert=\left|\frac {a}{e}\right|$$

For the series to be converging, $$\left\vert\frac {a}{e}\right\vert < 1 \implies-e <a<e$$

I don't think I have done this correctly but I don't see anything wrong with my steps.

• Can you fix the first line? It should be $a^n$, not $a_n$. – NickD Apr 10 '18 at 16:05
• so sorry, i haven't done this formatting before. – pinklemonade Apr 10 '18 at 16:06
• This certainly looks right to me. If you've made a mistake, I can't find it. To pick a nit, it's possible that the series converges at $e$ or $-e$ but not for $|a|>e$ – saulspatz Apr 10 '18 at 16:16
• @saulspatz for $|a|=e$ it also diverges by stirling – gimusi Apr 10 '18 at 17:23
• "For the series to be converging, $\left\vert\dfrac {a}{e}\right\vert < 1$" No, that's not correct. The other direction is correct: If the limit is $<1,$ then the series converges. – zhw. Apr 10 '18 at 19:06

## 1 Answer

Yes it is correct and by ratio test we can conclude that

• $|a|<e$ the series converges
• $|a|> e$ the series diverges

moreover since for $a=e$ by Stirling's approximation $n! \sim \sqrt{2 \pi n}\left(\frac{n}{e}\right)^n$

$$\dfrac {n!}{n^n}e^n\sim\sqrt{2 \pi n}\to\infty$$

for $|a|=e$ the series also diverges.