<s>[INST] 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?
 [/INST] 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 $n=0,1,2,3,\cdots$
$$T_0=\frac1{0!}$$
$$T_1=0+3\frac1{0!}+\frac1{1!}$$
$$T_2=\frac1{0!}+3\frac1{1!}+\frac1{2!}$$
$$T_3=\frac1{1!}+3\frac1{2!}+\frac1{3!}$$
$$\cdots$$
$$\text{So, }\sum_{0\le r<\infty}\frac{(n+1)^2}{n!}=e+3e+e$$
$$\text{So, }\sum_{1\le r<\infty}\frac{(n+1)^2}{n!}=e+3e+e-\frac{(0+1)^2}{0!}=5e-1$$
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