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Find all z such that the series ((-1)n z3n) / n!

Find all complex numbers z such that the series

    \[ \sum_{n=1}^{\infty} \frac{(-1)^n z^{3n}}{n!} \]

converges.


Consider,

    \[ \sum_{n=1}^{\infty} \left| \frac{(-1)^n z^{3n}}{n!} \right| = \sum_{n=1}^{\infty} \frac{z^{3n}}{n!}. \]

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*}

Hence, by the ratio test the series is absolutely convergent for all z \in \mathbb{C}.

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