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 $.
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.
(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.
