Counting factors: eg, how can I tell there are $n+1$ factors in $(n+\frac12) (n-\frac12)(n-\frac32)\cdots\frac32\cdot\frac12$?

Question

I am presented with the following problem:

Prove that: $$\left(n+\frac{1}{2}\right) \left(n-\frac{1}{2}\right)\left(n-\frac{3}{2}\right)\cdots\frac{3}{2}\cdot\frac{1}{2} = \frac{(2n+1)!}{2^{2n+1}n!}$$

The first steps I'm aware of are:

$$\frac{1}{2^{n+1}}(2n+1)(2n-1)\cdots 3\cdot1$$ I know from here I can then manipulate factorials and such to get the desired result.

My question is:

In a general sense for products like this, how can one tell that there are $n+1$ factors here (hence the power on the denominator), instead of $n$, or $n-1$?

I've tried to explain it to myself with the gamma function (i.e. that $\Gamma(n+1)$ has $n$ terms for integer n), but I'm not too certain if my logic is correct here.

Any help appreciated!

Answer 1

The product here can be rewritten as $$\left(n+ \frac{1}{2} \right)\left((n-1)+ \frac{1}{2} \right)\left((n-2)+ \frac{1}{2} \right)...\left(1+ \frac{1}{2} \right)\left(0+ \frac{1}{2} \right)$$

The number of factors is clearly the numbers of integers between $0$ and $n$, i.e. $n+1$.

Answer 2

If you have indices ranging from $a$ up to $b$ with a step of size $s$, the number of terms is $$ \overbrace{\ \ \frac{b-a}s\ \ }^{\substack{\text{number}\\\text{of steps,}\\\text{each with}\\\text{a term}}}+\overset{\substack{\text{last term}\\\downarrow\\[7pt]\,}}{1} $$

For example, $$ 3,7,11,15,19,23,27,31 $$ $s=4$, $a=3$, and $b=31$. The number of terms is $\frac{31-3}4+1=8$.

For this question, $s=1$, $a=\frac12$, and $b=n+\frac12$.