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:

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.