Find closed formula for $\prod_{i=2}^{n}(2i-3)$

Question

I am trying to find a close formula for $\prod\limits_{i=2}^{n}(2i-3)$.

So far I tried to take $\log$, so it would be calculating a sum which I will use $\exp$ later on.

I found it hard to calculate $\ln(1)+\ln(3)+\ln(5)+...+\ln(2n-3)$.

So I am stuck. It feels like its an easy exercise if you got some tricks I never studied.

Thanks a lot!

Answer 1

You do not need logarithms !

The given product calculates as $$1\cdot 3\cdot 5\cdots (2n-3)=\frac{(2n-2)!}{(n-1)!\cdot 2^{n-1}}$$

As Peter Foreman mentioned, this can also be written as the doublefactorial $$(2n-3)!!$$

Answer 2

I like to write this out as a recurrence relation: $$a_n=\prod_{i=2}^n(2i-3)=(2n-3)a_{n-1}=\frac{(2n-2)(2n-3)}{2(n-1)}a_{n-1}=\frac{(2n-2)!2^{n-1}(n-2)!}{(2n-4)!2^n(n-1)!}a_{n-1}$$ So $$\frac{2^n(n-1)!}{(2n-2)!}a_n=\frac{2^{n-1}(n-2)!}{(2n-4)!}a_{n-1}=\cdots=\frac{2^2(1!)}{2!}a_2=2$$ Which says $$a_n=\frac{(2n-2)!}{2^{n-1}(n-1)!}$$ Once you have the recurrence relation you can turn the coefficients into powers and gamma functions (or if you're lucky factorials) and rearrange into $f(n)=f(n-1)=\cdots=f(2)$ and then solve for $a_n$.

Answer 3

If you are not familiar with the double factorial but if you know the gamma function $$(2n-3)!!=\frac{2^{n-1} }{\sqrt{\pi }}\,\,\Gamma \left(n-\frac{1}{2}\right)$$