Exponential and logarithmic series: Find the sum of $2^2 + 3^2/ 2!+4^2/3!+...$ to infinity 
Find the sum of the following series:
  $ 2^2 + 3^2/2! + 4^2/3! + ...$ to infinity

The answer is given as $5e$ but I got it as $5e+1$
 $T_n = 1/(n-2)! +3/(n-1)! + 1/(n)! $ for $n \ge 2$
and $T_1+T_2 + ... $ to infinity = $4 + e + 3(e-1) +(e-2)$
$=5e-1$
Can you teell me which is correct and how?

You (after correction) : $\ \boxed{5\,e-1}$
\begin{align}
S(x):&=\sum_{k=1}^\infty \frac{x^{k+1}}{k!}&=x\,(e^x-1)\\
x\frac d{dx}S(x)&=\sum_{k=1}^\infty \frac{(k+1)\,x^{k+1}}{k!}&=x\,((x+1)e^x-1)\\
x\frac d{dx}\left(x\frac d{dx}S(x)\right)&=\sum_{k=1}^\infty \frac{(k+1)^2\,x^{k+1}}{k!}&=x\,((x^2+3x+1)e^x-1)\\\
\end{align}
Set $x=1$ to conclude.
