Radius of convergence of powerseries $\sum_{n=1}^\infty \frac{(-1)^n}{n!}z^n$

Question

$$ \begin{align} \sum_{n=1}^\infty \frac{(-1)^n}{n!}z^n \end{align} $$

Find the radius of convergence of this powerseries.

To determine the radius of convergence should I split it into two separate powerseries, one for $n=2k$ and one for $n=2k+1$ , or should I go straight and apply the ratio test or the root test?

Answer 1

With d'Alembert (I think, it's also known as "Ratio Test"):

$$\lim_{n\to\infty }\frac{\left|\frac{(-1)^{n+1}}{(n+1)!}\right|}{\left|\frac{(-1)^n}{n!}\right|}=\lim_{n\to\infty }\frac{n!}{(n+1)!}=\lim_{n\to\infty }\frac{1}{n+1}=0$$

Then the radius of convergence is $\mathcal R=\infty $.

Answer 2

You can ignore the $(-1)^n$ because the center of the disk of convergence is $0$, and $(-1)^n z^n=(-z)^n$, so the series without the alternating sign has the same radius of convergence. In other words, $z$ lies in the open disk of convergence if and only if $-z$ does too.