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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?

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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.

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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 $\frac{n}{e}$

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  • $\begingroup$ Thanks, so then it is $\frac{\infty}{3}$ and since it's bigger than 3, it diverges? $\endgroup$ – JJ Abrams Jun 24 at 22:14
  • $\begingroup$ Basically yes, although it is a bit sloppy to consider $\infty$ as a number, but in this context, I would say it is OK. $\endgroup$ – Peter Jun 24 at 22:20
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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 $.

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