# Proving formula for product of first n odd numbers

I have this formula which seems to work for the product of the first n odd numbers (I have tested it for all numbers from $1$ to $100$):

$$\prod_{i = 1}^{n} (2i - 1) = \frac{(2n)!}{2^{n} n!}$$

How can I prove that it holds (or find a counter-example)?

• This sounds like homework. Did you try induction? – cardinal Oct 2 '11 at 1:20
• @Pedro: Do you mean $$\left(\prod_{i=1}^n2i\right) - 1$$ (which is what you wrote) or $$\prod_{i=1}^n(2i-1)$$(which is what your title suggests)? – Arturo Magidin Oct 2 '11 at 1:25
• It is not, I stumbled upon this formula by accident and was wondering. The proof was fairly simple, I suppose I should have thought more about it – Pedro Oct 2 '11 at 1:28
• @ArturoMagidin: I meant the latter, sorry for the ambiguity – Pedro Oct 2 '11 at 1:30
• As a note: what you have is sometimes termed as the "double factorial". – J. M. is a poor mathematician Oct 2 '11 at 9:33

The idea is to "complete the factorials":

$$1\cdot 3 \cdot 5 \cdots (2n-1) = \frac{ 1 \cdot 2 \cdot 3 \cdot 4 \cdots (2n-1)\cdot (2n) }{2\cdot 4 \cdot 6 \cdots (2n)}$$

Now take out the factor of $2$ from each term in the denominator:

$$= \frac{ (2n)! }{2^n \left( 1\cdot 2 \cdot 3 \cdots n \right)} = \frac{(2n)!}{2^n n!}$$

A mathematician may object that there is a small gray area about what exactly happens between those ellipses, so for a completely rigorous proof one would take my post and incorporate it into a proof by induction.

For the induction argument, \begin{align*} \prod_{i=1}^{n+1}(2i-1)&=\left(\prod_{i=1}^n(2i-1)\right)(2n+1)\\ &= \frac{(2n)!(2n+1)}{2^n n!} \end{align*} by the induction hypothesis. Now multiply that last fraction by a carefully chosen expression of the form $\dfrac{a}a$ to get the desired result.

$$\Pi_{k=1}^n(2k-1) = \Pi_{k=1}^n(2k-1) \frac{\Pi_{k=1}^n(2k)}{\Pi_{k=1}^n(2k)} =\frac{\Pi_{k=1}^{2n}k}{2^n\Pi_{k=1}^n k} = \frac{(2n)!}{2^n n!}$$

• This is the same answer as the one given four years ago by Ragib, only with fewer explanations. It's good that you want to help by answering questions, but maybe you could consider answering some that haven't been answered yet? – mrf Aug 31 '15 at 20:42