Home » Blog » Prove some properties of binomial coefficients of a real number α

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}.

  1. For \alpha = -\frac{1}{2} show that

        \[ \binom{\alpha}{1} = -\frac{1}{2}, \quad \binom{\alpha}{2} = \frac{3}{8}, \quad \binom{\alpha}{3} = -\frac{5}{16}, \quad \binom{\alpha}{4} = \frac{35}{128}, \quad \binom{\alpha}{5} = -\frac{63}{256}. \]

  2. Consider the series whose terms are defined by

        \[ a_n = (-1)^n \binom{-\frac{1}{2}}{n}. \]

    Prove that

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


  1. 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*}

  2. 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

Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment):