Prove some properties of binomial coefficients of a real number α

The binomial coefficient $\binom{\alpha}{n}$ is defined by

$\binom{\alpha}{n} = \frac{\alpha(\alpha - 1)(\alpha - 2) \cdots (\alpha - n + 1)}{n!}$

where $\alpha \in \mathbb{R}$ and $n \in \mathbb{Z}_{\geq 0}$ .

For $\alpha = -\frac{1}{2}$ show that
$\binom{\alpha}{1} = -\frac{1}{2}, \quad \binom{\alpha}{2} = \frac{3}{8}, \quad \binom{\alpha}{3} = -\frac{5}{16}, \quad \binom{\alpha}{4} = \frac{35}{128}, \quad \binom{\alpha}{5} = -\frac{63}{256}.$
Consider the series whose terms are defined by
$a_n = (-1)^n \binom{-\frac{1}{2}}{n}.$

Prove that

$a_n > 0 \qquad \text{and} \qquad a_{n+1} < a_n.$

We compute each of the requested quantities using the given definition, where $\alpha = -\frac{1}{2}$ ,
$\begin{align*} \binom{\alpha}{1} &= \binom{-\frac{1}{2}}{1} = \frac{-\frac{1}{2}}{1} = -\frac{1}{2}. \\[10pt] \binom{\alpha}{2} &= \binom{-\frac{1}{2}}{2} = \left(-\frac{1}{2}\right)\left(-\frac{3}{\frac{2}{2}}\right) = \frac{3}{8}. \\[10pt] \binom{\alpha}{3} &= \binom{-\frac{1}{2}}{3} = \left( \frac{3}{8} \right) \left( \frac{-\frac{5}{2}}{3} \right) = -\frac{5}{16} \\[10pt] \binom{\alpha}{4} &= \binom{-\frac{1}{2}}{4} = \left(-\frac{5}{16} \right) \left( \frac{-\frac{7}{2}}{4} \right) = \frac{35}{128} \\[10pt] \binom{\alpha}{5} &= \binom{-\frac{1}{2}}{5} = \left( \frac{35}{128} \right) \left( \frac{-\frac{9}{2}}{5} \right) = -\frac{63}{256}. \end{align*}$
Proof. The proof is by induction. For $n = 1$ we have
$a_1 = (-1)^1 \binom{-\frac{1}{2}}{1} = (-1) \left( -\frac{1}{2} \right) = \frac{1}{2} > 0.$

For $n = 2$ we have

$a_2 = (-1)^2 \binom{-\frac{1}{2}}{2} = 1 \left( \frac{3}{8} \right) = \frac{3}{8}.$

Thus, for $n = 1$ we have

$a_n > 0 \qquad \text{and} \qquad a_2 < a_1.$

Hence, the statement holds for the case $n = 1$ . Assume then that the statement holds for $n =k \in \mathbb{Z}_{\geq 1}$ . Then,

$a_k > 0 \qquad \implies \qquad (-1)^k \binom{-\frac{1}{2}}{k} > 0.$

Furthermore, from the definition of $a_n$ we have

$\begin{align*} a_{k+1} &= a_k \cdot (-1) \left( \frac{-\frac{1}{2} - k + 1}{k} \right) \\[9pt] &= a_k \cdot \left( \frac{ \frac{1}{2} + k - 1}{k} \right) \\[9pt] &= a_k \cdot \left( 1 - \frac{1}{2k} \right). \end{align*}$

But, $1 - \frac{1}{2k} > 0$ for all positive integers $k$ , and $a_k > 0$ by the induction hypothesis. Thus, $a_{k+1} > 0$ . This gives us the first part of the statement, $a_n> 0$ for all positive integers $n$ . Next, since

$0 < \left( 1 - \frac{1}{2k} \right) < 1$

we have

$a_k \cdot \left( 1 - \frac{1}{2k} \right) < a_k \quad \implies \quad a_{k+1} < a_k.$

Hence, both statements are true for all positive integers $n. \qquad \blacksquare$