$\frac{(n/2)!}{n!} = \frac{1}{2^{n/2}(n-1)!!}$?

Question

I was working on a puzzle involving some rather complex probability when I arrived at two very distinct methods with very different ways of calculating the probability of solving the puzzle. The actual puzzle and how I got these probabilities isn't very important (they're also most likely wrong, I'm not very good with probability).

The functions I ended up with were $\frac{((n/2)!)^2}{n!}$ and $\frac{(n/2)!}{2^{n/2}(n-1)!!}$. To my surprise, when I plugged a few test values into Wolfram Alpha, I found that they were equivalent.

I divided both terms by $(n/2)!$ to get the equality $\frac{(n/2)!}{n!} = \frac{1}{2^{n/2}(n-1)!!}$, which I find totally unintuitive and rather interesting.

Is there already a proof/identity out there related to this equality? If not, is there a relatively simple proof using the definition of factorial that I'm not seeing? Is it even true?

Answer 1

Yes this is a well known relation.

It comes from the fact that by pulling a two from each term we have $$(2n)!!=2 \cdot 4 \cdot 6 \cdot...\cdot (2n)=2^n n!$$

Now combine this with $$(2n!!)[(2n+1)!!]=(2n+1)!$$ and you get your relation.

Answer 2

Let $n=2m$, and rewrite the identity as

$$2^mm!=\frac{(2m)!}{(2m-1)!!}\;.\tag{1}$$

Now notice both sides of $(1)$ are simply

$$\prod_{k=1}^m2k\;,$$

since cancellation removes the odd factors in the numerator on the righthand side.

Answer 3

Let $n = 2k$, then: $LHS = \dfrac{k!}{(2k)!}$, and $RHS = \dfrac{1}{2^k\cdot (2k-1)!!} = \dfrac{k!}{(2k)!!(2k-1)!!} = \dfrac{k!}{(2k)!} = LHS$. So it is a true statement.