$\sum_{n=1}^\infty \frac{1}{(n^2-1)!} - \sum_{n=1}^\infty \frac{1}{(7n+1)!}$ is almost $1+1/6$ I've recognized, that
$$\mathcal{S} = \sum_{n=1}^\infty \frac{1}{(n^2-1)!} - \sum_{n=1}^\infty \frac{1}{(7n+1)!} \approx 1.1666666666666666666657785992648796$$
which is almost $1+1/6$.
I think it is not a mathematical coincidence, but I'm not sure what are behind the scences.

Why is this value almost $1+1/6$?

In the first terms of the sequences $n^2-1$ [$A005563$] and $7n+1$ [$A016993$] the terms $8$ and $15$ coincide. Furthermore
$$\sum_{n=1}^\infty \frac{1}{(n^2-1)!} \approx 1.166691468254732970341437561639 $$
and
$$\sum_{n=1}^\infty \frac{1}{(7n+1)!} \approx 0.000024801588066303675658962374$$
Are there a family of this type of identities? Also would be nice to see the exact value of $1+1/6-\mathcal{S}$ which is approximately
$$
8.880674017870608390962304909410175868736976 \cdot 10^{-22}
$$

It follows because the first few terms are the same, and the terms in the series decrease to zero extremely quickly. Since$$\sum_{n=1}^{\infty}\frac{1}{(n^{2}-1)!}=1+\frac{1}{3!}+\frac{1}{8!}+\frac{1}{15!}+\frac{1}{24!}+\cdots,$$ we have that $$\left|\sum_{n=1}^{\infty}\frac{1}{(n^{2}-1)!}-\left(1+\frac{1}{3!}+\frac{1}{8!}+\frac{1}{15!}\right)\right|\leq\frac{2}{24!}$$ and since
$$\sum_{n=1}^{\infty}\frac{1}{(7n+1)!}=\frac{1}{8!}+\frac{1}{15!}+\frac{1}{22!}+\cdots$$ we have that 
$$\left|\sum_{n=1}^{\infty}\frac{1}{(7n+1)!}-\left(\frac{1}{8!}+\frac{1}{15!}\right)\right|\leq\frac{2}{22!},$$ and so the difference of the sums is $$\left|\sum_{n=1}^{\infty}\frac{1}{(n^{2}-1)!}-\sum_{n=1}^{\infty}\frac{1}{(7n+1)!}-\left(1+\frac{1}{6}\right)\right|\leq\frac{3}{22!} .$$ Now, if we were a bit more careful we could replace $3$ with $1+\epsilon$ where $\epsilon>0$ is a small positive constant around $1/100$. Notice that the error is almost exactly $1/22!$ since $$\frac{1}{22!}=8.8967913924505732867\dots \times 10^{-22}.$$
