**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](http://oeis.org/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](http://oeis.org/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](http://en.wikipedia.org/wiki/Oppermann%27s_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](http://mathworld.wolfram.com/ContinuedFraction.html),


$$
\frac{\sum \_{k=0}^{\infty } \frac{1}{(k+2) k!}}{\sum \_{m=0}^{\infty } \left(\sum \_{k=m}^{\infty } \frac{1}{(k+2) k!}\right)}=\frac{1}{1+\frac{2}{2+\frac{3}{3+\frac{4}{4+\frac{5}{5+\frac{6}{6+\frac{7}{7+\frac{8}{8+\frac{9}{9+\frac{10}{10+11}}}}}}}}}}
$$


**Another interesting pattern for the series:**  

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



$\sum _{k=0}^{\infty } \frac{1}{(k+m) k!} \equiv 1$ for $m=2$
A way to get this, and also to understand the behavior for other values would be like so (though I do not know if this is not overly indirect):


Recall that 
$$
e^x = \sum\_{k=0}^{\infty} \frac{x^k}{k!}
$$
so 
$$
x^{m-1}e^x = \sum\_{k=0}^{\infty} \frac{x^{k+m-1}}{k!}
$$
Now 'integerate', then 
$$
F(x) = \sum\_{k=0}^{\infty} \frac{x^{k+m}}{(k+m)k!}
$$
where $F$ is some antiderivative of $x^{m-1}e^x$.


For $m=2$ one has the antiderivatives $e^x(x-1) +c$.
Setting $x=0$ one finds that $F(x) = e^x(x-1) +1$.
Setting $x=1$ one finds $1=\sum\_{k=0}^{\infty} \frac{1}{(k+2)k!}$.
Now subtract the term for $k=0$, which is $1/2$ to get your result. 


(Not sure this is on-topic, but it is weekend and I was bored. Sorry in advance, to those how might mind.)


