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

## Theorem

Consider the series:

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

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

## Proof

First we show that this series converges when $n \in \Z_{\ge 0}$.

Consider the coefficients:

$1, \paren {1 - \dfrac 1 {1!} }, \paren {1 - \dfrac 1 {1!} + \dfrac 1 {2!} }, \ldots$

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

Starting from the $\paren {n + 1}$th term, there is a factor of $\paren {n - n}$.

In this case, all subsequent terms of the expansion equal $0$.

Therefore the series converges.

$\Box$

Then we show that this series diverges for $n \notin \Z_{\ge 0}$.

We do this via Divergence Test.

That is, we show that $\displaystyle \lim_{k \mathop \to \infty} \frac { {!k} \, n^{\underline k} } {k!} \ne 0$.

Aiming for a contradiction, suppose $\displaystyle \lim_{k \mathop \to \infty} \frac { {!k} \, n^{\underline k} } {k!}$ exists.

By Power Series Expansion for Exponential Function, we have that $\displaystyle \lim_{k \mathop \to \infty} \frac {k!} {!k} = e$.

Hence $\displaystyle \lim_{k \mathop \to \infty} n^{\underline k}$ would exist as well.

For $k > \floor {n + 2}$:

 $\ds n - k$ $<$ $\ds -2 + n + 2 - \floor {n + 2}$ $\ds$ $\le$ $\ds -2$ $\ds \leadsto \ \$ $\ds \size {n^{\underline k} }$ $=$ $\ds \size {n^{\underline {k - 1} } } \size {n - k}$ $\ds$ $>$ $\ds 2 \size {n^{\underline {k - 1} } }$

Since $n - k \ne 0$, $\size {n^{\underline k} }$ increases without bound.

Hence its limit does not exist, a contradiction.

Therefore $\displaystyle \lim_{k \mathop \to \infty} \frac { {!k} \, n^{\underline k} } {k!}$ does not exist for $n \notin \Z_{\ge 0}$.

By Divergence Test, our series diverges.

$\blacksquare$