Show that the denominator of \[ \binom{1/2}{n} \] is a power of \(2\) for every integer \(n\).
Solution
Use the standard integer-index convention \[ \binom{\alpha}{n}=0 \qquad(n<0), \] and for \(n\geq0\) use the usual generalized binomial formula.
1. If \(n<0\), then the reduced denominator of\[ \binom{1/2}{n} \]is \(1\), hence a power of \(2\).
By the stated convention the binomial coefficient is \(0\), whose reduced rational form is \(0/1\). Since \[ 1=2^0, \] the claim holds.
2. If \(n=0\), then the reduced denominator is again \(1\).
One has \[ \binom{1/2}{0}=1. \]
3. Suppose \(n\geq1\). Then\[ \binom{1/2}{n} = (-1)^{n-1} \frac{(2n-3)!!}{2^n n!}. \]
Expanding the generalized binomial coefficient gives \[ \begin{aligned} \binom{1/2}{n} &= \frac{ (1/2)(1/2-1)(1/2-2)\cdots(1/2-n+1) }{n!}\\ &= (-1)^{n-1} \frac{1\cdot3\cdot5\cdots(2n-3)}{2^n n!}. \end{aligned} \] For \(n=1\), the empty odd product is interpreted as \((-1)!!=1\), so the same formula applies.
4. For every \(n\geq1\),\[ C_{n-1} \coloneqq \frac1n\binom{2n-2}{n-1} \]is an integer.
The identity \[ \frac1n\binom{2n-2}{n-1} = \binom{2n-2}{n-1} - \binom{2n-2}{n-2} \] expresses \(C_{n-1}\) as a difference of two integers.
5. For \(n\geq1\),\[ \binom{1/2}{n} = (-1)^{n-1} \frac{C_{n-1}}{2^{2n-1}}. \]
The double-factorial identity \[ (2n-2)! = 2^{n-1}(n-1)!(2n-3)!! \] gives \[ \begin{aligned} C_{n-1} &= \frac{(2n-2)!}{n!(n-1)!}\\ &= 2^{n-1} \frac{(2n-3)!!}{n!}. \end{aligned} \] Hence \[ \frac{(2n-3)!!}{n!} = \frac{C_{n-1}}{2^{n-1}}. \] Substitute this into step 3.
6. For every \(n\geq1\), the reduced denominator of\[ \binom{1/2}{n} \]is a power of \(2\).
By steps 4 and 5, the number has the form \[ \frac{m}{2^{2n-1}} \] with \(m\in\ZZ\). Reducing this fraction can only cancel factors of \(2\) from the denominator. Therefore the reduced denominator is \[ 2^r \] for some integer \(r\geq0\).
7. The claim holds for every integer \(n\).
Steps 1 and 2 handle \(n\leq0\), and step 6 handles \(n\geq1\).
8. Q.E.D.
Step 7 is the required conclusion.