# $\sum _{k=0}^{\infty } \frac{1}{(k+m) k!} \equiv 1$ for $m=2$

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

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This question is oddly written. Why even mention $n$? Why link to the OEIS for the decimal expansion of $0.5$? Why sum from $k=1$ instead of from $k=0$? – Douglas Zare Apr 7 '13 at 0:54
Taking $m=2$ in $1 / ((k+m)k!)$ gives $1/((k+2)k!) = 1/(k+1)! - 1/(k+2)!$. Therefore the sum telescopes: $$\sum_{k=1}^\infty \frac1{(k+2)k!} = \sum_{k=1}^\infty \frac1{(k+1)!} - \frac1{(k+2)!} = \left(\frac1{2!}-\frac1{3!}\right) + \left(\frac1{3!}-\frac1{4!}\right) + \left(\frac1{4!}-\frac1{5!}\right) + \cdots = \frac12,$$ QED. – Noam D. Elkies Apr 7 '13 at 6:02
@Fred Kline: Why wouldn't $0$ count as the square root of $0$? – Douglas Zare Apr 7 '13 at 16:30
Sorry, I think I misunderstood your question as asking for the general phenomenon specified after the dash. For this specific series, the information regarding the name is more or less in my answer (the commment). More specifically, on the wiki-page of the incomplete Gamma function I mentioned note the function $\gamma(s,x)$, the lower incomplete Gamma function. Further down (between ref [20] and [21]) as part of a long formula a series for it is given. If $x= -1$ and $s=m$ you get exactly your series except for the factor of $(-1)^m$ at the start. Or still differently your series is... – user9072 Apr 9 '13 at 14:23
...could be obtained by considering (the series expansion) of $\gamma(s,x)x^{-s}$ at $x=-1$ or likely still other expressions using the functions there. Except that 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). – user9072 Apr 9 '13 at 14:28

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.)

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+1, for info I can use. Anyone who has an opinion on Q2, leave a comment. Thanks. – Fred Kline Apr 7 '13 at 0:08
You are welcome. I am not completely sure what you intend with Q2 and/or what is missing that this answers it, perhaps this helps. I assume $m$ positive integer. An antiderivative is $(-1)^{m+1} \Gamma(m, -x)$ where the $\Gamma$ is incomplete Gamma func (this is almost tautological with the defintiion of that function) see en.wikipedia.org/wiki/Incomplete_gamma_function Now, $\Gamma(m,0)= (m-1)!$ from this you get that the $c$ and it is rational. And $\Gamma(m,-1) = (m-1)! e \sum_{i=0}^{m-1} (-1)^i /i!$ (only $m=2$ the sum vanishes). [Better recheck the details though.] – user9072 Apr 7 '13 at 0:32
Sorry I did not check back what Q2 was, I thought it was the question if m=2 is somehow a unique value. I do not know OEIS well enough to voice an opinion. – user9072 Apr 7 '13 at 0:51
Very infrequently, I add interesting series that I have found to OEIS entries. Their entry for $1/2$ has only two. Would this interesting enough to be included? I like the fact about the sum vanishing. Since I'm working with the square root, I am hoping the $1/2$ can be shown as $n^{\frac{1}{2}}=k$. – Fred Kline Apr 7 '13 at 0:52
I guess our comments 'crossed.' Please see the preceeding one. – user9072 Apr 7 '13 at 0:54

Another way is to write $1/(k+2)$ as $1/(k+1) - 1/((k+1)(k+2))$.

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+1 not sure why the later comment saying essentially the same thing already has three up-votes and this had none. – user9072 Apr 7 '13 at 21:44
@quid: to be honest I didn't check the comments above when I wrote the answer either:) – John Jiang Apr 7 '13 at 22:03

In the comments, quid showed that $\sum_{k=0}^\infty \frac{1}{(k+m)k!} = (-1)^m\bigg((m-1)! - !(m-1)e \bigg)$ where $!a = a! \sum_{k=0}^a \frac{(-1)^k}{k!}$ is the number of derangements in the symmetric group on $a$ objects. For example, $\sum_{k=0}^\infty \frac{1}{(k+10)k!} = 9! - !9 e = 362880-133496e$.

This explains why $m=2$ is the only case which is rational. The magnitude of the coefficient of $e$ is the number of derangements in $S_{m-1}$, and the only case where the number of derangements is $0$ is in $S_1$. While $S_0 \cong S_1$, a crucial difference between the trivial group $S_0$ and the trivial group $S_1$ is that the identity in $S_0$ is a derangement, and the corresponding sum for $m=1$ is $-1+e$.

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+1, Now it's making sense to me, I think. This shows that the $k+2$ segments as I have described them, have no derangements. That is good. Thank-you. – Fred Kline Apr 16 '13 at 4:43
I apologize for arguing against your suggestion of a starting index of zero. Since I revised my approach, everything is falling into place. – Fred Kline Apr 16 '13 at 4:47
And thanks for the edit. – Fred Kline Apr 16 '13 at 5:01