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

\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?

• Note that you should be able to ascertain the radius of convergence if you can recognize what function this is the Taylor series of. That'll let you confirm the answer you get via convergence tests. – Semiclassical Aug 20 '14 at 19:40
• I suspect you mean $(-1)^n$ since $(-1^n) = -1$ for all $n$. – Fly by Night Aug 20 '14 at 19:42
• sorry that's what I meant indeed – helplessKirk Aug 20 '14 at 19:52
• @Semiclassical I believe $f(z)=\frac{1-e^z}{e^z}$ – helplessKirk Aug 20 '14 at 20:07
• Quite so, though you can write it more simply as $e^{-z}-1$. So all that matters is the radius of convergence of an exponential function... – Semiclassical Aug 20 '14 at 20:10

$$\lim_{n\to\infty }\frac{\left|\frac{(-1)^{n+1}}{(n+1)!}\right|}{\left|\frac{(-1)^n}{n!}\right|}=\lim_{n\to\infty }\frac{n!}{(n+1)!}=\lim_{n\to\infty }\frac{1}{n+1}=0$$
Then the radius of convergence is $\mathcal R=\infty$.
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.