Solution
= Solution
Let $a_n=2^n/n!$. The <ratio test> gives $a_{n+1}/a_n=2/(n+1)\to0<1$, so \b[the series converges absolutely]. In fact the <exponential series> identifies its value as
$$
\sum_{n=1}^{\infty}\frac{2^n}{n!}=e^2-1,
$$
with the $n=0$ term subtracted.