Factorial as Product of Consecutive Factorials/Lemma 2

Theorem

Let $n \in \N$.

Then $\paren {2 n - 2}! \, \paren {2 n - 1}! > \paren {3 n - 1}!$ for all $n \ge 7$.

Proof

We prove the result by induction on $n$.

Basis for the Induction

The case $n = 7$ is verified by direct calculation:

$12! \times 13! > 20!$

This is the basis for the induction.

Induction Hypothesis

We suppose for some $k \ge 7$, we have:

$\paren {2 k - 2}! \, \paren {2 k - 1}! > \paren {3 k - 1}!$

This is our induction hypothesis.

Induction Step

This is our induction step:

\(\ds \paren {2 k}! \, \paren {2 k + 1}!\)

\(>\)

\(\ds \paren {2 k} \paren {2 k - 1} \paren {2 k + 1} \paren {2 k} \paren {3 k - 1}!\)

Induction Hypothesis

\(\ds \)

\(>\)

\(\ds k \paren {2 k - 1} \paren {2 k + 1} \paren {3 k} \paren {3 k - 1}!\)

because $4 k > 3 k$

\(\ds \)

\(>\)

\(\ds 4 \paren {2 k - 1} \paren {2 k + 1} \paren {3 k} \paren {3 k - 1}!\)

because $k > 4$

\(\ds \)

\(=\)

\(\ds \paren {4 k - 2} \paren {4 k + 2} \paren {3 k} \paren {3 k - 1}!\)

\(\ds \)

\(\ge\)

\(\ds \paren {3 k + 5} \paren {3 k + 9} \paren {3 k} \paren {3 k - 1}!\)

because $k \ge 7$

\(\ds \)

\(>\)

\(\ds \paren {3 k + 1} \paren {3 k + 2} \paren {3 k} \paren {3 k - 1}!\)

\(\ds \)

\(=\)

\(\ds \paren {3 k + 2}!\)

Definition of Factorial

The result follows by the Principle of Mathematical Induction.

$\blacksquare$