Simplifying a sequence formula I wonder if this formula can be simplified (so that will be no sum symbol):
$$a_n = n!\sum _{k=0} ^{n}{1 \over k!} $$
If you have any ideas utalizing generating functions or Z transform, post please
 A: Using the Taylor polynomial centered at $0$ for the function $\mathrm{e}^{x}$,
we have that
$$
\mathrm{e}^{x}=\sum_{k=0}^{n}\frac{x^{k}}{k!}+\frac{x^{n+1}\mathrm{e}^{\theta
x}}{(n+1)!}\,,\quad\theta=\theta(x)\in(0,1)
$$
meanig that, for $x:=1$,
$$
n!\sum_{k=0}^{n}\frac{1}{k!}=n!\left(  \mathrm{e}-\frac{\mathrm{e}^{\theta}
}{(n+1)!}\right)  =n!\mathrm{e}-\frac{\mathrm{e}^{\theta}}{n+1}
$$
so
$$
n!\sum_{k=0}^{n}\frac{1}{k!}\in\left(  n!\mathrm{e}-\frac{\mathrm{e}}
{n+1},n!\mathrm{e}-\frac{\mathrm{1}}{n+1}\right)  \text{.}
$$
If $n\geq1$, then the interval $\left(  n!\mathrm{e}-\dfrac{\mathrm{e}}
{n+1},n!\mathrm{e}-\dfrac{\mathrm{1}}{n+1}\right)  $ has length less then $1$,
hence $n!\displaystyle\sum_{k=0}^{n}\dfrac{1}{k!}$ is the only integer
that lies inside this interval.
For any real number $x$, denote by $[x]$ the integer part of $x$, i.e., the
largest integer that does not exceed $x$; hence $[x]$ is the only integer that
lies in the interval $(x-1,x]$. This means that $[n!\mathrm{e}]\in\left(  n!\mathrm{e}
-1\mathbf{,}n!\mathrm{e}\right]  $. If $n\geq2$, then
$$
\left(  n!\mathrm{e}-\frac{\mathrm{e}}{n+1},n!\mathrm{e}-\frac{\mathrm{1}
}{n+1}\right)  \subset\left(  n!\mathrm{e}-1\mathbf{,}n!\mathrm{e}\right]
$$
hence $\left(  n!\mathrm{e}-\frac{\mathrm{e}}{n+1},n!\mathrm{e}-\frac
{\mathrm{1}}{n+1}\right)  $ contains at most one integer, and this must be
$[n!\mathrm{e}]$. So
$$
n!\sum_{k=0}^{n}\frac{1}{k!}=[n!\mathrm{e}]
$$
if $n\geq2$.
A: $$1+1!x\left({1\over 0!}+{1\over 1!}\right)+2!x^2\left({1\over 0!}+{1\over 1!}+{1\over 2!}\right)+\cdots=(1+x+x^2+\cdots)+(x+2x^2+3x^3+\cdots)+(1\cdot 2x^2+2\cdot 3x^3+\cdots)+\cdots=\frac{1}{1-x}+x\frac{d}{dx}\left(\frac{1}{1-x}\right)+{x^2\over 2!}\frac{d^2}{dx^2}\left(\frac{1}{1-x}\right)+\cdots=\left(1+x\frac{d}{dx}+{x^2\over 2!}\frac{d^2}{dx^2}+\cdots\right)\left(\frac{1}{1-x}\right)=e^{x\frac{d}{dx}}\left(\frac{1}{1-x}\right)$$
So $$n!\sum_{k=0}^{n}{{1\over k!}}=[x^n]e^{x\frac{d}{dx}}\left(\frac{1}{1-x}\right)$$ where $[x^n]f(x)$ is the coefficient of $x^n$ in the power series of $f(x)$. 
Just to clarify, I have no idea what any of that means :)
