Answer
$\Sigma\frac{1}{k!}$
$\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\frac{1}{(k+1)!}\times\frac{k!}{1}|=\lim\limits_{n \to \infty}|\frac{1}{k+1}|=\frac{1}{\infty}=0$. Convergent.
Work Step by Step
Using the ratio test we find that the limit equals 0. Since 0<1 it can be stated that the series is absolutely convergent.