Factorial Greater than Cube for n Greater than 5

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Theorem

Let $n \in \Z$ be an integer such that $n > 5$.


Then $n! > n^3$.


Proof 1

We note that:

\(\ds 1!\) \(=\) \(\ds 1\)
\(\ds \) \(=\) \(\ds 1^3\)
\(\ds 2!\) \(=\) \(\ds 2\)
\(\ds \) \(<\) \(\ds 8\)
\(\ds \) \(=\) \(\ds 2^3\)
\(\ds 3!\) \(=\) \(\ds 6\)
\(\ds \) \(<\) \(\ds 27\)
\(\ds \) \(=\) \(\ds 3^3\)
\(\ds 4!\) \(=\) \(\ds 24\)
\(\ds \) \(<\) \(\ds 64\)
\(\ds \) \(=\) \(\ds 4^3\)
\(\ds 5!\) \(=\) \(\ds 120\)
\(\ds \) \(<\) \(\ds 125\)
\(\ds \) \(=\) \(\ds 5^3\)


The proof then proceeds by induction.


For all $n \in \Z_{\ge 6}$, let $\map P n$ be the proposition:

$n! > n^3$


Basis for the Induction

$\map P 6$ is the case:

\(\ds 6!\) \(=\) \(\ds 720\)
\(\ds \) \(>\) \(\ds 216\)
\(\ds \) \(=\) \(\ds 6^3\)

Thus $\map P 6$ is seen to hold.


This is the basis for the induction.


Induction Hypothesis

Now it needs to be shown that if $\map P k$ is true, where $k \ge 6$, then it logically follows that $\map P {k + 1}$ is true.


So this is the induction hypothesis:

$k! > k^3$


from which it is to be shown that:

$\paren {k + 1}! > \paren {k + 1}^3$


Induction Step

This is the induction step:

\(\ds \paren {k + 1}!\) \(=\) \(\ds \paren {k + 1} \times k!\)
\(\ds \) \(>\) \(\ds \paren {k + 1} \times k^3\) Induction Hypothesis
\(\ds \) \(=\) \(\ds \paren {k - 3} \times k^3 + 3 \times k^3 + k^3\)
\(\ds \) \(>\) \(\ds 1 + 3 \times k + 3 \times k^2 + k^3\)
\(\ds \) \(=\) \(\ds \paren {k + 1}^3\)

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.


Therefore:

$\forall n \in \Z_{>5}: n! > n^3$

$\blacksquare$


Proof 2

For $n > 5$, notice that the following inequalities hold:

$2 \paren {n - 1} = 2 n - 2 > n + 5 - 2 > n$
$3 \paren {n - 2} = 3 n - 6 > n + 10 - 6 > n$

And thus:

\(\ds n!\) \(\ge\) \(\ds n \paren {n - 1} \paren {n - 2} \paren {3!}\)
\(\ds \) \(=\) \(\ds n \paren {2 \paren {n - 1} } \paren {3 \paren {n - 2} }\)
\(\ds \) \(>\) \(\ds n \paren n \paren n\)
\(\ds \) \(=\) \(\ds n^3\)

$\blacksquare$


Sources

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