# Divergence of series with n factorial

I need to find the convergence or divergence of $$\sum_{n=1}^\infty \frac{n!}{3^n}$$ The first thing I did was to take the nth root. So I got: $$\lim_{n\rightarrow\infty}\big(\frac{n!}{3^n}\big)^\frac{1}{n} = \lim_{n\rightarrow\infty} \frac{(n!)^\frac{1}{n}}{3}$$

And since $$(n!)^\frac{1}{n}$$ converges to 1, the limit I got is $$\frac{1}{3}$$, which is less than 1 so it would converge. BUT, the series diverges. Was I wrong in the calculations or what other convergence test can I use? I tried using the aspect ratio and the limit gave me $$\infty$$, so naturally it's bigger than 1, therefore it diverges as the series actually does. Did I applied the nth root wrongly or is it an special case in which it can't be applied?

Another approach: Note that for $$n>3,$$
$$\frac{n!}{3^n}=\frac{n}{3}\cdots \frac{3}{3}\frac{2}{3}\frac{1}{3}>\frac{2}{9}.$$ Thus the $$n$$th term does not appoach $$0,$$ hence the series diverges.
The flaw is that $$(n!)^{1/n}$$ for $$n\rightarrow \infty$$ does not converge to $$1$$, but diverges to $$\infty$$
To see this, use Stirlings approximation giving that $$(n!)^{1/n}$$ is asymptotically $$\frac{n}{e}$$
• Thanks, so then it is $\frac{\infty}{3}$ and since it's bigger than 3, it diverges? – JJ Abrams Jun 24 at 22:14
• Basically yes, although it is a bit sloppy to consider $\infty$ as a number, but in this context, I would say it is OK. – Peter Jun 24 at 22:20
By the ratio test, the series diverges, since $$\lim_{n\to\infty}\vert\dfrac {a_{n+1}}{a_n}\vert=\lim_{n\to\infty}\dfrac {n+1}3=\infty$$.