Summation involving subfactorial function Inspired by this post:
Does the following series converge; if so, to what value does it converge? $$ \sum_{n = 2}^\infty \left|\frac{!n \cdot e}{n!} - 1\right|$$ I am looking for a closed form for the second question.
Note: !n denotes the subfactorial, also known as the number of derangments within $ \mathrm{S}_n $. 
 A: Some calculations: We have
$$
\sum_{n=2}^\infty\left|\frac{!n\cdot e}{n!}-1\right|=\sum_{n=2}^\infty\left|\sum_{k=0}^\infty e\frac{(-1)^k}{k!}-1\right|=\sum_{n=2}^\infty\left|\frac{\Gamma(n+1,-1)}{\Gamma(n+1)}-1\right|.
$$
We also have
$$
\frac{\Gamma(n+1,-1)}{\Gamma(n+1)}=\frac{\Gamma(n+1)-\gamma(n+1,-1)}{\Gamma(n+1)}=1+\Gamma(n+1)^{-1}\int_{-1}^0e^{-t}t^n\mathrm{d}t.
$$
EDIT: Oops, I had a factorial too many. Removed.
A: (This is a corrected version)
I get that the sum converges,
but do not know the value.
Since $!n = n!\sum_{k=0}^n\frac{(-1)^k}{k!}$,
$\frac{!n \cdot e}{n!}
= e \sum_{k=0}^n\frac{(-1)^k}{k!}
= e \big(\frac1{e}-\sum_{k=n+1}^{\infty}\frac{(-1)^k}{k!}\big)
= 1-e\sum_{k=n+1}^{\infty}\frac{(-1)^k}{k!}
$.
Starting as Carl Najafi did,
$\begin{align*}
\sum_{n=2}^{\infty}\left|\frac{!n\cdot e}{n!}-1\right|
&=\sum_{n=2}^{\infty}\left|\big(1-e\sum_{k=n+1}^{\infty}\frac{(-1)^k}{k!}\big)-1\big)\right|\\
&=e\sum_{n=2}^{\infty}\left|\sum_{k=n+1}^{\infty}\frac{(-1)^{k+1}}{k!}\right|\\
&=e\sum_{n=2}^{\infty}\left|(-1)^{n}\sum_{k=n+1}^{\infty}\frac{(-1)^{k-n+1}}{k!}\right|\\
&=e\sum_{n=2}^{\infty}\left|(-1)^{n}\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{(n+k)!}\right|\\
&=e\sum_{n=2}^{\infty}\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{(n+k)!}\\
\end{align*}
$
The inner sum
$S(n) 
= \sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{(n+k)!}
$
is an alternating series
with terms decreasing in absolute value to $0$,
so its sum is bounded
by any two consecutive sums.
In particular,
taking the first two sums,
$\frac1{(n+1)!}
> S(n) > \frac1{(n+1)!} - \frac1{(n+2)!}
> \frac1{2(n+1)!}
$.
Therefore
$\sum_{n=2}^{\infty}\left|\frac{!n\cdot e}{n!}-1\right|
< \sum_{n=2}^{\infty}\frac1{(n+1)!}
$
which easily converges.
My original answer
actually looked at
$\sum_{n=2}^{\infty}\left|!n\cdot e-n!\right|
=\sum_{n=2}^{\infty}\left|n! S(n)\right|
$.
Since 
$\frac1{n+1}
> |n!S(n)| > \frac1{n+1} - \frac1{(n+1)(n+2)}
\ge \frac1{2(n+1)}
$,
this sum diverges, but can be shown to converge
if the absolute value signs are removed,
since the resulting series is alternating
and decreasing.
