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 $$x_n = \frac {x^n} {n!}$$.

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 $$\forall n > N: \frac {x^n} {n!} \le \frac {x^{N-1}} {\left({N-1}\right)!} \left({\frac {x} {N}}\right)^{n - N + 1}$$.

Note that as $$N > x$$, $$\frac x N < 1$$.

Also, $$m > n \implies \frac x m < \frac x n$$.

Thus:

$$ $$ $$ $$

As $$\frac x N < 1$$, it follows from Power of a Number Less Than One that $$\left({\frac x N}\right)^n \to 0$$ as $$n \to \infty$$.

For a given $$x$$ and $$N$$, $$\frac {x^{N-1}} {\left({N-1}\right)!} \left({\frac x N}\right)^{1 - N}$$ is constant.

Thus by the Combination Theorem for Sequences, $$\frac {x^{N-1}} {\left({N-1}\right)!} \left({\frac x N}\right)^{1 - N} \left({\frac x N}\right)^n \to 0$$ as $$n \to \infty$$.

As (from above) $$\frac {x^n} {n!} \le \frac {x^{N-1}} {\left({N-1}\right)!} \left({\frac x N}\right)^{1 - N} \left({\frac x N}\right)^n$$, the result follows from the Squeeze Theorem.