I changed the title and added revisions and left the original untouched
For this post, $k$ is defined to be the square root of some $n\geq k^{2}$. Out of curiousity, I took the sum of one of the factorials in the denominator of the binomial theorem; $$\sum _{k=1}^{\infty } \frac{1}{k!} \equiv e-1$$ OEIS A091131
Because I need to show that only the contiguous non-overlapping sequences of size $k$ up to $k^{2}+2k$ are valid for my purpose, I took the same sum with the denominator multiplied by $k+2$: $$\sum _{k=1}^{\infty } \frac{1}{(k+m) k!} \equiv \frac{1}{2}\text{ for $m=2$ }$$ OEIS A020761
This is not a sum that I expected.
When $m\neq2$ the convergence returns alternating values like $\frac{1}{k}(-x+y e)$ and $\frac{1}{k}(x^{\prime}-y^{\prime} e)$, so $\frac{1}{2}$ seems to be the only value constructed out of integers.
Two questions:
$1)$ Is there a proof technique that can use this specific convergence to show that $k+2$ is the natural limit to my sequences? And that those specific non-overlapping sequences are the only ones that apply?
$2)$ Is this convergence interesting enough to put into OEIS?
I need some hints for my next step.
Edit
Q1 is answered. I have enough info to keep me going for a few months.
Q2: if you look at the OEIS entries for constants like $\pi$ and $e$, you will see dozens of identities. The entry for $\frac{1}{2}$ has only two identities. I feel it should have many more. But, just because I find this series interesting, doesn't mean others do, therefore, the question.
My motivation is to prove Oppermann's conjecture. Thanks for the great answers and comments, and your patience.
Revised
Original post revised to use $k=0$ as starting index. And we show an example of the underlying pattern.
$ e= \sum_{k=0}^{\infty} 1/k!\textit{ Revised }$
$ e-1= \sum_{k=0}^{\infty} 1/((k+m)k!)\text{ for }m=1$
$ 1= \sum_{k=0}^{\infty} 1/((k+m)k!)\text{ for }m=2$
$\sum_{k=0}^{\infty} 1/((k+m)k!)\not \in \textbf{Q} \text{ for }m>2$
Example of underlying pattern for (say) $k=3$:
$(1, 2, 3), (4, 5, 6), (7, 8, 9), (10, 11, 12), (13, 14, 15)$
$(1, 2, 3), (1, 2, 3), (1, 2, 3), (1, 2, 3), (1, 2, 3)$
$(1, 2, 3), (2, 1, 2), (1, 2, 3), (2, 1, 2), (1, 2, 3)$
Top: Number line partitioned into $k+2$ non-overlapping ordered lists
Middle: Equivalence classes $n-1 \mod k +1$
Bottom: Least divisors. $1= p_{x}$
What is it about these patterns that causes the convergence result for $m=2$ to be $\in \textbf{Q}$?
Coda
Removed the identities as not quite in step. Below I show the summand of my function on left, the summand of an 'instep' identity, and a variation of the identity.
$$\frac{1}{(k+2)k!} \equiv \frac{1}{(k+1)!+k!} \equiv \frac{1}{\Gamma(k+2)+k!}$$
So, $\frac{1}{(k+2)k!}$ sums two consecutive factorials. Why?
New This ratio equals $(e-1)^{-1}$ as shown here,
$$ \frac{\sum _{k=0}^{\infty } \frac{1}{(k+2) k!}}{\sum _{m=0}^{\infty } \left(\sum _{k=m}^{\infty } \frac{1}{(k+2) k!}\right)}=\frac{1}{1+\frac{2}{2+\frac{3}{3+\frac{4}{4+\frac{5}{5+\frac{6}{6+\frac{7}{7+\frac{8}{8+\frac{9}{9+\frac{10}{10+11}}}}}}}}}} $$
Another interesting pattern for the series:
$$
11_2,22_3,33_4,44_5,55_6,66_7,77_8,88_9,99_{10},\text{AA}_{11},\text{BB}_{12},\text{CC}_{13}{}{}{}
$$