Find all complex numbers 
such that the series
![\[ \sum_{n=1}^{\infty} \frac{(-1)^n z^{3n}}{n!} \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-0c8662438b216e65fc0d777de12c50f3_l3.png "Rendered by QuickLaTeX.com")
converges.
Consider,
![\[ \sum_{n=1}^{\infty} \left| \frac{(-1)^n z^{3n}}{n!} \right| = \sum_{n=1}^{\infty} \frac{z^{3n}}{n!}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-6b03a4195fdedcc1ee6897a3f3f33979_l3.png "Rendered by QuickLaTeX.com")
Then, by the ratio test we have
![\begin{align*} \lim_{n \to \infty} \frac{a_{n+1}}{a_n} &= \lim_{n \to \infty} \left( \frac{z^{3n+3}}{(n+1)!} \right) \left( \frac{n!}{z^{3n}} \right) \\[9pt] &= \lim_{n \to \infty} \frac{z^3}{n+1} \\[9pt] &= 0 \qquad \text{for all }z \in \mathbb{C}. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-ea0cd94598050d4e8c9fde0f8abc17b7_l3.png "Rendered by QuickLaTeX.com")
Hence, by the ratio test the series is absolutely convergent for all 
.
Find all complex numbers such that the series
converges.
Consider,
Then, by the ratio test we have
Hence, by the ratio test the series is absolutely convergent for all .
it should be abs(z) since z is complex