Factorial as Sum of Series of Subfactorial by Falling Factorial over Factorial/Condition for Convergence

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Theorem

Consider the series:

\(\displaystyle n!\) \(=\) \(\displaystyle \sum_{k \mathop \ge 0} \dfrac { {!k} \, n^{\underline k} } {k!}\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac { !0 \times n^{\underline 0} } {0!} + \dfrac { {!1} \times n^{\underline 1} } {1!} + \dfrac { {!2} \times n^{\underline 2} } {2!} + \dfrac { {!3} \times n^{\underline 3} } {3!} + \cdots\)
\(\displaystyle \) \(=\) \(\displaystyle 1 + \left({1 - \dfrac 1 {1 !} }\right) n + \left({1 - \dfrac 1 {1 !} + \dfrac 1 {2 !} }\right) n \left({n - 1}\right) + \left({1 - \dfrac 1 {1 !} + \dfrac 1 {2 !} - \dfrac 1 {3 !} }\right) n \left({n - 1}\right) \left({n - 2}\right) + \cdots\)


This converges only when $n \in \Z_{\ge 0}$, that is, when $n$ is a non-negative integer.


Proof

Consider the coefficients:

$1, \left({1 - \dfrac 1 {1!} }\right), \left({1 - \dfrac 1 {1!} + \dfrac 1 {2!} }\right), \ldots$


By Power Series Expansion for Exponential Function, they converge to $\dfrac 1 e$.

Thus none of the terms ever reaches $0$ except when there is a factor of $\left({n - n}\right)$.

In this case, all subsequent terms of the expansion equal $0$ and indeed, the sequence converges.

$\blacksquare$



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