Formal power series manipulations and a closed formula for $\sum_{n\geq 0}{\frac{n^2+4n+5}{n!}}$ I'm reading a book on generating functions, and in their formal power series section they define:
If $f \overset{ops}{\leftrightarrow} \left \{ a_n \right \}_{0}^{\infty}$, and $P$ is a polynomial, then
$P(xD)f \overset{ops}{\leftrightarrow} \left \{ P(n)a_{n} \right \}_{n\geq0}$  
I'm having issues understanding the notation being used. The example they give is:
Find a closed formula for the sum of the series $\sum_{n\geq 0}{\frac{n^2+4n+5}{n!}}$  
They continue with:
$\left \{ (xD)^2+4(xD)+5 \right \}e^x = \left \{ x^2+x \right \}e^x+4xe^x+5e^x$  
I can't seem to figure out where the stand alone "x" came from in $x^2+x$  
 A: For $a_0+a_1n+a_2n^2+a_3n^3+\cdots$ in the numerator,  where $a_i$s are given constants, 
we need to express it as $b_0+b_1n+b_2n(n-1)+b_3n(n-1)(n-2)+\cdots$ where $b_i$s are arbitrary constants 
$a_0+a_1n+a_2n^2+a_3n^3+\cdots=b_0+b_1n+b_2n(n-1)+b_3n(n-1)(n-2)+\cdots$
Set $n=0,1,2\cdots$ in the above identity to express $b_i$s in terms of $a_i$s
$$\implies S=\sum_{n=0}^\infty\frac{a_0+a_1n+a_2n^2+a_3n^3+\cdots}{n!}$$
$$=\sum_{n=0}^\infty\frac{b_0+b_1n+b_2n(n-1)+b_3n(n-1)(n-2)+\cdots}{n!}$$
$$=b_0\sum_{n=0}^\infty\frac1{n!}+b_1\sum_{n=0}^\infty\frac1{(n-1)!}+b_2\sum_{n=0}^\infty\frac1{(n-2)!}+b_3\sum_{n=0}^\infty\frac1{(n-2)!}+\cdots$$
As $\dfrac1{r!}=0$ for integer $r<0,$
$$S=b_0\sum_{n=0}^\infty\frac1{n!}+b_1\sum_{n=1}^\infty\frac1{(n-1)!}+b_2\sum_{n=2}^\infty\frac1{(n-2)!}+b_3\sum_{n=3}^\infty\frac1{(n-2)!}+\cdots$$
$$=e(b_0+b_1+b_2+b_3+\cdots)$$
A: You know that:
$\begin{align*}
  e^z
    &= \sum_{n \ge 0} \frac{z^n}{n!}
\end{align*}$
Now, if $\langle a_n \rangle$ is any sequence, we can write:
$\begin{align*}
   A(z)
     &= \sum_{n \ge 0} a_n \frac{z^n}{n!} \\
   z \frac{d}{d z} A(z)
     &= \sum_{n \ge 0} n a_n \frac{z^n}{n!}
\end{align*}$
So we can write:
$\begin{align*}
   \sum_{n \ge 0} (n^2 +4 n + 5) \frac{z^n}{n!}
     &= z \frac{d}{d z} \left( z \frac{d}{d z} \right) e^z
         + 4 z \frac{d}{d z} e^z
         + 5 e^z \\
     &= (z^2 + 5 z + 5) e^z
\end{align*}$
The value asked for is just the above at $z = 1$, i.e. $11 e$.
