Determine if the following series converges or diverges and justify your decision.
![\[ \sum_{n=1}^{\infty} \frac{n!}{3^n}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-9acd8ef98212e10702df6c3587aac563_l3.png "Rendered by QuickLaTeX.com")
The series diverges by the ratio test since
![\begin{align*} \lim_{n \to \infty} \frac{a_{n+1}}{a_n} &= \lim_{n \to \infty} \left( \frac{(n+1)!}{3^{n+1}} \right) \left( \frac{3^n}{n!} \right) \\[9pt] &= \lim_{n \to \infty} \frac{n+1}{3} \\[9pt] &= +\infty > 1. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-52b985f26dd0e64b415e0da086060b7e_l3.png "Rendered by QuickLaTeX.com")
Hence, the series diverges.
Determine if the following series converges or diverges and justify your decision.
The series diverges by the ratio test since
Hence, the series diverges.
Infinity is not a number to compare it with one, you can use the limit, however, to establish the fact that the sequence a_n is increasing for n>N and thus cannot approach zero (since a_n>0 for all n)