**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:**

$$
11_2,22_3,33_4,44_5,55_6,66_7,77_8,88_9,99_{10},\text{AA}_{11},\text{BB}_{12},\text{CC}_{13}{}{}{}
$$

QED. – Noam D. Elkies Apr 7 '13 at 6:02Exceptthat these series typically start at 0 not at 1, so you have to account for the constant term. But basically what you are considering are special values of the incomplete Gamma function(s). – quid Apr 9 '13 at 14:28