What explains these numerators being the same

combinatoricsfactorialsequences-and-series

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?

Best Answer

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 $k$.

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