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Find all z such that the series (z-1)n / (n+2)! converges

Find all complex numbers z such that the series

    \[ \sum_{n=0}^{\infty} \frac{(z-1)^n}{(n+2)!} \]

converges.


Consider,

    \begin{align*}  \lim_{n \to \infty} \frac{a_{n+1}}{a_n} &= \lim_{n \to \infty} \left( \frac{(z-1)^{n+1}}{(n+3)!} \right) \left( \frac{(n+2)!}{(z-1)^n} \right) \\[9pt]  &= \lim_{n \to \infty} \frac{z-1}{n+3} \\[9pt]  &= 0 \qquad \text{for all }z \in \mathbb{C}. \end{align*}

Hence, the series converges for all complex z.

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