Factorial Greater than Cube for n Greater than 5

From ProofWiki
Jump to navigation Jump to search

Theorem

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


Then $n! > n^3$.


Proof

We note that:

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


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:

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

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:

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

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


Therefore:

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

$\blacksquare$


Sources