I got this interesting sum which seems to involve values of the derangement problem: $$ \sum _{n=0}^{\infty } \frac{1}{(2 n+2) (2 n)!}=\frac{e-1}{e}=1-\frac{1}{e},$$ where $1-\frac{1}{e}$ is the probability that some man gets his own hat back [OEIS A068996].
Breaking down the intermediate expression to: $\frac{1}{e} \times (e-1),$ we have the probability that no man gets his own hat back [OEIS A068985], times the Engel Extension [OEIS A091131] for the natural numbers.
Is there any deep meaning?
On
Let's first derive the sum.
$$f(x) = \sum_{n=0}^{\infty} \frac{x^{2 n+2}}{(2 n+2) (2 n)!} $$
$$\implies f'(x) = x \sum_{n=0}^{\infty} \frac{x^{2 n}}{(2 n)!} = x \cosh{x}$$
$$\implies f(x) = 1- \cosh{x}+x\sinh{x} $$
Thus,
$$\sum_{n=0}^{\infty} \frac{1}{(2 n+2) (2 n)!} = 1 - (\cosh{1} - \sinh{1}) = 1-\frac1{e} $$
Now, to answer your question, I do not know anything about deep meaning concerning the Engel's extension. However, I do find it interesting that this sum is equal to
$$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n!} $$
METHODOLOGY $1$:
One approach relies on the identity
$$\begin{align} \sum_{n=0}^\infty \frac{(-1)^n}{n!}&=\sum_{n,\text{even}}^\infty\frac{1}{n!}-\sum_{n,\text{odd}}^\infty\frac{1}{n!}\\\\ &=1+\sum_{n=1}^\infty\frac{1}{(2n)!}-\sum_{n=1}^\infty\frac{1}{(2n-1)!} \tag 1 \end{align}$$
Then, we can write
$$\begin{align} \sum_{n=0}^\infty\frac{1}{(2n+2)(2n)!}&=\sum_{n=0}^\infty\frac{2n+1}{(2n+2)!}\\\\ &=\sum_{n=1}^\infty\frac{2n-1}{(2n)!}\\\\ &=\sum_{n=1}^\infty \left(\frac{1}{(2n-1)!}-\frac{1}{(2n)!}\right) \tag 2\\\\ &=1-\sum_{n=0}^\infty\frac{(-1)^n}{n!} \tag 3\\\\ &=1-\frac1e \end{align}$$
as was to be shown!
Note in going from $(2)$ to $(3)$, we made use of the identity given in $(1)$
METHODOLOGY $2$:
$$\begin{align} \sum_{n=0}^\infty\frac{1}{(2n+2)(2n)!}&=\sum_{n=0}^\infty\frac{1}{(2n)!}\int_0^1x^{2n+1}\,dx\\\\ &=\int_0^1 \sum_{n=0}^\infty\frac{x^{2n+1}}{(2n)!}\,dx\\\\ &=\int_0^1x\cosh(x)\,dx\\\\ &=1-\frac1e \end{align}$$
as expected!