<s>[INST] Sum $\sum_{n=1}^\infty \frac{n^2}{(n+2)!}$ Problem is to find sum
$$\frac{1}{3!}+\frac{4}{4!}+\frac{9}{5!}\cdots$$
What I knew doesn't apply on this problem
Some series are telescoping, some types are solvable using binomial , both look useless here
Binomial gives
$$n (n-1) (n-2)x^3=1$$
 $$n (n-1)( n-2)(n-3)x^4=4$$
What approach to use here?
 [/INST] $$\sum\limits_{n = 0}^{ + \infty } {\frac{{n^2 }}{{\left( {n + 2} \right)!}}}  = \sum\limits_{n = 1}^{ + \infty } {\frac{{n^2 }}{{\left( {n + 2} \right)!}}}  =  - 5 + 2\sum\limits_{n = 0}^{ + \infty } {\frac{{\left( {n - 1} \right)^2 }}{{n!}}}  =  - 5 + 2e
$$
$$
\begin{array}{l}
 s = \sum\limits_{n = 0}^{ + \infty } {\frac{{n^2 }}{{\left( {n + 2} \right)!}}}  = \sum\limits_{n = 2}^{ + \infty } {\frac{{\left( {n - 2} \right)^2 }}{{n!}}}  =  - \frac{{\left( {0 - 2} \right)^2 }}{{0!}} - \frac{{\left( {1 - 2} \right)^2 }}{{1!}} + \sum\limits_{n = 0}^{ + \infty } {\frac{{\left( {n - 2} \right)^2 }}{{n!}}}  \\ 
 s =  - 5 + \sum\limits_{n = 0}^{ + \infty } {\frac{{\left( {n - 2} \right)^2 }}{{n!}}}  =  - 5 + \sum\limits_{n = 0}^{ + \infty } {\frac{{\left( {n - 2} \right)^2 }}{{n!}}}  \\ 
 s =  - 5 + \sum\limits_{n = 0}^{ + \infty } {\frac{{n^2  - 4n + 4}}{{n!}}}  =  - 5 + \sum\limits_{n = 0}^{ + \infty } {\frac{{n^2 }}{{n!}}}  - \sum\limits_{n = 0}^{ + \infty } {\frac{{4n}}{{n!}}}  + \sum\limits_{n = 0}^{ + \infty } {\frac{4}{{n!}}}  \\ 
 s =  - 5 + \frac{{0^2 }}{{0!}} + \sum\limits_{n = 1}^{ + \infty } {\frac{{n^2 }}{{n!}}}  - \frac{{4\left( 0 \right)}}{{0!}} - \sum\limits_{n = 1}^{ + \infty } {\frac{{4n}}{{n!}}}  + \sum\limits_{n = 0}^{ + \infty } {\frac{4}{{n!}}}  \\ 
 s =  - 5 + \sum\limits_{n = 1}^{ + \infty } {\frac{n}{{\left( {n - 1} \right)!}}}  - \sum\limits_{n = 1}^{ + \infty } {\frac{4}{{\left( {n - 1} \right)!}}}  + \sum\limits_{n = 0}^{ + \infty } {\frac{4}{{n!}}} \quad ;\quad \left( {\sum\limits_{n = 1}^{ + \infty } {\frac{4}{{\left( {n - 1} \right)!}}}  = \sum\limits_{n = 0}^{ + \infty } {\frac{4}{{n!}}} } \right) \\ 
 s =  - 5 + \sum\limits_{n = 1}^{ + \infty } {\frac{n}{{\left( {n - 1} \right)!}}}  =  - 5 + \sum\limits_{n = 1}^{ + \infty } {\frac{{n - 1 + 1}}{{\left( {n - 1} \right)!}}}  =  - 5 + \sum\limits_{n = 1}^{ + \infty } {\frac{{n - 1}}{{\left( {n - 1} \right)!}}}  + \sum\limits_{n = 1}^{ + \infty } {\frac{1}{{\left( {n - 1} \right)!}}}  \\ 
 s =  - 5 + 2e \\ 
 \end{array}
$$
</s>