Factorial as Product of Consecutive Factorials/Lemma 1
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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 | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {2 n}! \paren {\prod_{k \mathop = 1}^n k} \paren {\prod_{k \mathop = n + 1}^{2 n - 1} k}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {2 n}! \paren {\prod_{k \mathop = 0}^{n - 1} \paren {k + 1} } \paren {\prod_{k \mathop = 1}^{n - 1} \paren {2 n - k} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {2 n}! \prod_{k \mathop = 1}^{n - 1} \paren {k + 1} \paren {2 n - k}\) | $0 + 1 = 1$ | |||||||||||
\(\ds \) | \(>\) | \(\ds \paren {2 n}! \prod_{k \mathop = 1}^{n - 1} \paren {2 n + k}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {3 n - 1}!\) |
$\blacksquare$