Solution
= Solution
For the positive terms $u_n=2^n/n!$, the consecutive ratio is $u_{n+1}/u_n=2/(n+1)\to0$, so the <ratio test> proves convergence. The <exponential series> gives $e^2=\sum_{n=0}^{\infty}2^n/n!$. Removing the $n=0$ term yields
$$
\boxed{\sum_{n=1}^{\infty}\frac{2^n}{n!}=e^2-1.}
$$