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In figuring out the number of terms of different order in the adjugate from this question, I stumbled upon this fact:

The numerators for the alternating sum of fractional factorials (https://oeis.org/A053557)

\begin{align} E_{k} = \left\{ 1, 0, \tfrac{1}{2}, \tfrac{1}{3}, \tfrac{3}{8}, \tfrac{11}{30}, \tfrac{53}{144}, \tfrac{103}{280}, \tfrac{2119}{5760}, ... \right\} \end{align} where

\begin{align} E_{k} =\sum _{m=0}^{k}\frac{( -1)^{m}}{m!} \end{align}

... matches the numerators for its cumulative sum (skipping $E_0,E_1$ though, I guess):

\begin{align} F_{k} =\sum _{m=0}^{k} E_{m} \end{align}

\begin{align} F_{k} = \left\{ 1, 1, \tfrac{3}{2}, \tfrac{11}{6}, \tfrac{53}{24}, \tfrac{103}{40}, \tfrac{2119}{720}, \tfrac{16687}{5040}, \tfrac{16481}{4480}, ... \right\} \end{align}

What explains this?

julianiacoponi
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    I notice that if we divide $E_k$ by the corresponding $F_j$ with the same numerator, we get $\frac12, \frac13, \frac14, \frac 15, \frac 16, \frac17$. Is that suggestive? – MJD May 15 '24 at 18:03
  • As suggested by another question of mine (related to the adjugate question mentioned), these numbers are clearly closely related to the harmonic numbers for sure (which isn't surprising given the similarity of their arithmetic form). But yeah, that's a very cool extra thing you've spotted! @MJD – julianiacoponi May 16 '24 at 08:48

1 Answers1

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We want to prove that $F_{k-2}$ and $E_k$ have the same numerator for $k \geq 3$. If we prove these two facts for $k \geq 3$ we are done:

  1. $F_{k-2} = kE_k$.
  2. The denominator of $E_k$ after simplifications is a multiple of $k$.

Lets prove them.

1) \begin{align*} F_{k-2} &= \sum_{m=0}^{k-2}\sum_{j=0}^{m} \frac{(-1)^j}{j!}\\ &= \sum_{m=0}^{k-2} (k-1-m) \frac{(-1)^m}{m!}\\ &= (k-1)\sum_{m=0}^{k-2}\frac{(-1)^m}{m!} - \sum_{m=0}^{k-2}m\frac{(-1)^m}{m!}\\ &= (k-1)\sum_{m=0}^{k-2}\frac{(-1)^m}{m!} + \sum_{m=1}^{k-2}\frac{(-1)^{m-1}}{(m-1)!}\\ &= (k-1)\sum_{m=0}^{k-2}\frac{(-1)^m}{m!} + \sum_{m=0}^{k-3}\frac{(-1)^m}{m!}\\ &= k\sum_{m=0}^{k-2}\frac{(-1)^m}{m!} - \frac{(-1)^{k-2}}{(k-2)!}\\ &= k\sum_{m=0}^{k-2}\frac{(-1)^m}{m!} + (k-1)\frac{(-1)^{k-1}}{(k-1)!}\\ &= k\sum_{m=0}^{k-1}\frac{(-1)^m}{m!} -\frac{(-1)^{k-1}}{(k-1)!}\\ &= k\sum_{m=0}^{k-1}\frac{(-1)^m}{m!} +k\frac{(-1)^{k}}{k!}\\ &= k\sum_{m=0}^{k}\frac{(-1)^m}{m!}\\ &= kE_k \end{align*}

2)

\begin{align*} E_k &= \sum_{m=0}^{k}\frac{(-1)^m}{m!}\\ &=\frac{\sum_{m=0}^k (-1)^m \frac{k!}{m!}}{k!} \end{align*}

The denominator is clearly a multiple of $k$ and all the terms in the numerator are multiples of $k$ except for the case when $m=k$. Therefore, the numerator is not divisible by $k$ and the denominator of $E_k$ after simplifications will still be a multiple of $k$.

Villa
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  • Thanks so much for your detailed answer / derivation @Villa ! I am new to all this combinatorical maths so it's gratifying to learn stuff from each step of your derivation of 1).

    Could you clarify a few things?

    a) Why ignore $k=2,3$, just because they're trivial?

    b) Can you provide more explanation on why just points 1. and 2. are sufficient for the proof?

     – julianiacoponi May 17 '24 at 11:10
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    I am glad to help. It should be $k \geq 3$, I just edited. In my derivation I use $\sum_{m=0}^{k-3}$, and this has no sense if $k<3$. For your second question, if the denominator of $E_k$ is a multiple of $k$, then when you multiply $kE_k$, the $k$ gets simplified with the denominator, leaving the numerator unchanged. I recommend taking particular examples of the series to see this more clearly. – Villa May 17 '24 at 15:10
  • I can now see how the denominator 'absorbs' the factor of $k$, yes! And then of course, this explains @MJD 's comment on how the ratio equals the harmonic numbers too, since $E_k / F_{k-2} = 1/k$ ... bravo! And we ignore the $k=2$ case simply because... it is so trivial? – julianiacoponi May 20 '24 at 12:23