Power over Factorial

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Theorem

Let $x \in \R: x > 0$ be a positive real number.

Let $\sequence {x_n}$ be the sequence in $\R$ defined as $x_n = \dfrac {x^n} {n!}$.

Then $\sequence {x_n}$ 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:

$\forall n > N: \dfrac {x^n} {n!} \le \dfrac {x^{N - 1} } {\paren {N - 1}!} \paren {\dfrac x N}^{n - N + 1}$

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

Also:

$m > n \implies \dfrac x m < \dfrac x n$

Thus:

\(\displaystyle \frac {x^n} {n!}\) \(=\) \(\displaystyle \frac x 1 \frac x 2 \cdots \frac x {N - 1} \frac x N \frac x {N + 1} \cdots \frac x n\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {x^{N - 1} } {\paren {N - 1}!} \frac x N \frac x {N + 1} \cdots \frac x n\)
\(\displaystyle \) \(\le\) \(\displaystyle \frac {x^{N - 1} } {\paren {N - 1}!} \paren {\frac x N}^{n - N + 1}\) as $\dfrac x {N + 1}, \dfrac x {N + 2}, \ldots, \dfrac x n < \dfrac x N$
\(\displaystyle \) \(=\) \(\displaystyle \frac {x^{N - 1} } {\paren {N - 1}!} \paren {\frac x N}^{1 - N} \paren {\frac x N}^n\)


As $\dfrac x N < 1$, it follows from Sequence of Powers of Number less than One that $\paren {\dfrac x N}^n \to 0$ as $n \to \infty$.

For a given $x$ and $N$, $\dfrac {x^{N - 1} } {\paren {N - 1}!} \paren {\dfrac x N}^{1 - N}$ is constant.

Thus by the Multiple Rule for Real Sequences:

$\dfrac {x^{N - 1} } {\paren {N - 1}!} \paren {\dfrac x N}^{1 - N} \paren {\dfrac x N}^n \to 0$ as $n \to \infty$

As (from above):

$\dfrac {x^n} {n!} \le \dfrac {x^{N - 1} } {\paren {N - 1}!} \paren {\dfrac x N}^{1 - N} \paren {\dfrac x N}^n$

the result follows from the Squeeze Theorem for Real Sequences.

$\blacksquare$


Sources