<s>[INST] Find series power of $F(x) =e^{-x}x^{2}$ i need help for this problem; find  a power series for $F(x) \text{=}e^{-x}x^{2} $ and derivate and prove this expression  $$ \sum \limits^{\infty }_{n=1}\frac{(-2)^{n+1}(n+2)}{n!} =\text{4}$$
 [/INST] For $f(x) = x^{2} \, e^{-x}$ 
\begin{align}
f(x) = x^{2} \, \sum_{n=0}^{\infty} \frac{(-1)^{n} \, x^{n}}{n!} = \sum_{n=0}^{\infty} \frac{(-1)^{n} \, x^{n+2}}{n!}.
\end{align}
Differentiation leads to
\begin{align}
f'(x) = (2 - x) \, x \, e^{-x} = \sum_{n=0}^{\infty} \frac{(-1)^{n} \, (n+2) \, x^{n+1}}{n!}
\end{align}
Now let $x = 2$ to obtain
\begin{align}
\sum_{n=0}^{\infty} \frac{(-2)^{n} \, (n+2)}{n!} = 0
\end{align}
or, by shifting the first term to the right hand side, 
\begin{align}
\sum_{n=1}^{\infty} \frac{(-2)^{n} \, (n+2)}{n!} = -2.
\end{align}
Multiply both sides by $-2$ to obtain
\begin{align}
\sum_{n=1}^{\infty} \frac{(-2)^{n+1} \, (n+2)}{n!} = 4.
\end{align}
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