Solution
= Solution
For $a_n=n^2/n!$, the <ratio test> gives
$$
\frac{a_{n+1}}{a_n}=\frac{(n+1)^2}{n^2(n+1)}=\frac{n+1}{n^2}\longrightarrow0<1.
$$
Thus the series \b[converges absolutely]. As a check, $n^2=n(n-1)+n$ and the exponential <power series> give the sum $e+e=2e$.