Factorial as Product of Consecutive Factorials/Lemma 1

Theorem

Let $n \in \N$.

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

Proof

Let $n, k \in \N_{> 0}$.

Suppose $n > 1$ and $n > k$.

We show that $\paren {k + 1} \paren {2 n - k} > 2 n + k$.

For $k = 1$:

$2 \paren {2 n - 1} = 4 n - 2 \ge 2 n + 2 > 2 n + 1$

For $k > 1$:

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

\(=\)

\(\ds 2 n k + 2 n - k^2 - k\)

\(\ds \)

\(>\)

\(\ds 2 k^2 + 2 n - k^2 - k\)

because $n > k$

\(\ds \)

\(=\)

\(\ds k^2 + 2 n - k\)

\(\ds \)

\(>\)

\(\ds 2 k + 2 n - k\)

because $k \ge 2$

\(\ds \)

\(=\)

\(\ds 2 n + k\)

Therefore we have:

\(\ds \paren {2 n - 1}! \, \paren {2 n}!\)

\(=\)

\(\ds \paren {2 n}! \prod_{k \mathop = 1}^{2 n - 1} k\)

Definition of Factorial

\(\ds \)

\(=\)

\(\ds \paren {2 n}! \paren {\prod_{k \mathop = 1}^n k} \paren {\prod_{k \mathop = n + 1}^{2 n - 1} k}\)

\(\ds \)

\(=\)

\(\ds \paren {2 n}! \paren {\prod_{k \mathop = 0}^{n - 1} \paren {k + 1} } \paren {\prod_{k \mathop = 1}^{n - 1} \paren {2 n - k} }\)

\(\ds \)

\(=\)

\(\ds \paren {2 n}! \prod_{k \mathop = 1}^{n - 1} \paren {k + 1} \paren {2 n - k}\)

$0 + 1 = 1$

\(\ds \)

\(>\)

\(\ds \paren {2 n}! \prod_{k \mathop = 1}^{n - 1} \paren {2 n + k}\)

\(\ds \)

\(=\)

\(\ds \paren {3 n - 1}!\)

$\blacksquare$