Why is $1 \times 3 \times 5 \times \cdots \times (2k-3) = \frac{(2k-2)!}{2^{(k-1)}(k-1)!}$

Question

In order to find out the Catalan numbers from their generating function you have to evaluate the product above. Here is what I thought: \begin{align*} 1 \times 3 \times 5 \times...\times (2k-3) &= \frac{1 \times 2 \times 3 \times 4 \times \cdots \times (2k - 3)}{2 \times 4 \times \cdots \times (2k - 2)} \\ &= \frac{(2k - 3)!}{2^{k - 1}(k - 1)!} \\ \end{align*}

But a Mathematica session quickly proved me wrong, instead the result in the title is true. Why is that true, what mistake did I make?

Answer 1

Note that all even numbers in the numerator cancel out with the denominator. But in your calculation, $2k-2$ from denominator doesn't cancel out.
The correct way to arrive at the answer is:

$$\begin{align}1 \times 3 \times 5 \cdots \times (2k-3) &= 1 \times \dfrac{2}{2} \times 3 \times \dfrac{4}{4} \times \cdots \times (2k-3) \times \dfrac{2k-2}{2k-2} \\&= \dfrac{1\times 2\times \cdots \times (2k-3) \times (2k-2)}{2\times 4 \times \cdots \times (2k-2)} \\&= \dfrac{(2k-2)!}{2^{k-1}(k-1)!}\end{align}$$

Answer 2

You have missed to multiply the numerator by $2n-3+1=2n-2$

Answer 3

For example, for the denominator $$2\times 4\times 6\times 8=2\cdot 1\times 2\cdot 2\times 2\cdot 3\times 2\cdot 4=2\times 2\times 2\times 2\cdot 1\times 2\times 3\times 4=2^4\cdot 4!$$ I let you adapt this hint to your exercise.