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?
 A: 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.
A: The answer is $5e-1$. There might be some confusion with the initial values of the sum.
By letting $T_n = \frac{1}{(n-2)!} +  \frac{3}{(n-1)!} +  \frac{1}{n!}$, as you did:
$S = 2^2 + T_2 + T_3 + \dots = 4 + (\frac{1}{0!} +  \frac{3}{1!} +  \frac{1}{2!}) + (\frac{1}{1!} +  \frac{3}{2!} +  \frac{1}{3!}) + (\frac{1}{2!} +  \frac{3}{3!} +  \frac{1}{4!}) + \dots$
Then we can take the $\frac{1}{0!}$, $\frac{3}{1!}$, and $\frac{1}{1!}$ out of the brackets, giving us $S=4 + \frac{1}{0!} + \frac{3}{1!} + \frac{1}{1!} + 5(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\dots) = 4+1+3+1+5(e-2) = 5e-1$.
A: So, $$T_n=\frac{(n+1)^2}{n!}$$
Let $(n+1)^2=n(n-1)+Bn+C$
$\implies n^2+2n+1=n^2+n(B-1)+C$
$\implies B-1=2,B=3, C=1$
So, $$T_n=\frac{(n+1)^2}{n!}=\frac1{(n-2)!}+3\frac1{(n-1)!}+\frac1{n!}$$
Putting $n=0,1,2,3,\cdots$
$$T_0=\frac1{0!}$$
$$T_1=0+3\frac1{0!}+\frac1{1!}$$
$$T_2=\frac1{0!}+3\frac1{1!}+\frac1{2!}$$
$$T_3=\frac1{1!}+3\frac1{2!}+\frac1{3!}$$
$$\cdots$$
$$\text{So, }\sum_{0\le r<\infty}\frac{(n+1)^2}{n!}=e+3e+e$$
$$\text{So, }\sum_{1\le r<\infty}\frac{(n+1)^2}{n!}=e+3e+e-\frac{(0+1)^2}{0!}=5e-1$$
