Power over Factorial
Theorem
Let $x \in \R: x > 0$ be a positive real number.
Let $\left \langle {x_n} \right \rangle$ be the sequence in $\R$ defined as $x_n = \dfrac {x^n} {n!}$.
Then $\left \langle {x_n} \right \rangle$ converges to zero.
Proof
We need to show that $x_n \to 0$ as $n \to \infty$.
Let $N \in \N$ be the smallest natural number which satisfies $N > x$.
(From the Archimedean Principle, such an $N$ always exists.)
First we show that:
- $\displaystyle \forall n > N: \frac {x^n} {n!} \le \frac {x^{N-1}} {\left({N-1}\right)!} \left({\frac {x} {N}}\right)^{n - N + 1}$
Note that as $N > x$, $\dfrac x N < 1$.
Also, $\displaystyle m > n \implies \frac x m < \frac x n$.
Thus:
| \(\displaystyle \) | \(\displaystyle \frac {x^n} {n!}\) | \(=\) | \(\displaystyle \frac x 1 \frac x 2 \cdots \frac x {N-1} \frac x N \frac x {N+1} \cdots \frac x n\) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \frac {x^{N-1} } {\left({N-1}\right)!} \frac x N \frac x {N+1} \cdots \frac x n\) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\le\) | \(\displaystyle \frac {x^{N-1} } {\left({N-1}\right)!} \left({\frac x N}\right)^{n - N + 1}\) | \(\displaystyle \) | as $\displaystyle \frac x {N+1}, \frac x {N+2}, \ldots, \frac x n < \frac x N$ | ||
| \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \frac {x^{N-1} } {\left({N-1}\right)!} \left({\frac x N}\right)^{1 - N} \left({\frac x N}\right)^n\) | \(\displaystyle \) |
As $\dfrac x N < 1$, it follows from Power of a Number Less Than One that $\left({\dfrac x N}\right)^n \to 0$ as $n \to \infty$.
For a given $x$ and $N$, $\displaystyle \frac {x^{N-1}} {\left({N-1}\right)!} \left({\frac x N}\right)^{1 - N}$ is constant.
Thus by the Combination Theorem for Sequences:
- $\displaystyle \frac {x^{N-1}} {\left({N-1}\right)!} \left({\frac x N}\right)^{1 - N} \left({\frac x N}\right)^n \to 0$ as $n \to \infty$
As (from above):
- $\displaystyle \frac {x^n} {n!} \le \frac {x^{N-1}} {\left({N-1}\right)!} \left({\frac x N}\right)^{1 - N} \left({\frac x N}\right)^n$
the result follows from the Squeeze Theorem.
Sources
- K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 4.20 \ (4)$