Applying the definition of the binomial coefficient I can't figure out how to simplify the following expression:


I want to find:


I'm stuck at the beginning. $$a_{n}=\frac{-\frac{1}{2}!}{n!\left ( n-\frac{1}{2} \right )!}$$

Thank you so much!

  • $\begingroup$ Write it out as $\frac{-\frac{1}{2}\left(-\frac{1}{2}-1 \right)\left(-\frac{1}{2}-2\right)\cdots \left(-\frac{1}{2}-n+1 \right)}{n!}$. Take out a minus from each of the $n$ factors in the numerator. To simplify further multiply the fraction by $\frac{2^n}{2^n}$ and multiply in a factor of $2$ to each of the terms in the numerator. $\endgroup$ – Winther Dec 12 '16 at 0:15
  • 2
    $\begingroup$ Possible duplicate of Showing $\binom{2n}{n} = (-4)^n \binom{-1/2}{n}$. See also Generalized binomial coefficient $\endgroup$ – Winther Dec 12 '16 at 0:20
  • $\begingroup$ @Winther: While the accepted answer to the question that you cite also answers this question, this is obviously not a duplicate. $\endgroup$ – Brian M. Scott Dec 12 '16 at 0:27
  • $\begingroup$ @BrianM.Scott It's obviously not not obviously not. Anyway, I don't have strong feelings about it. If non of those answers does satisfy OPs needs and he/she wants answers that explains other aspects of this computation (like the title hints towards) then OP should just say so and I'll remove the vote. $\endgroup$ – Winther Dec 12 '16 at 0:35

Here we use the following definition of the binomial coefficient \begin{align*} \binom{\alpha}{n}=\frac{\alpha(\alpha-1)\cdots(\alpha-n+1)}{n!} \qquad\qquad\alpha\in\mathbb{C},\quad n\in \mathbb{N}\tag{1} \end{align*} It is also convenient to use double factorials for a more compact notation \begin{align*} (2n)!! &= (2n)(2n-2)\cdots 4\cdot2\qquad\qquad\qquad n\in\mathbb{N}\\ (2n-1)!! &= (2n-1)(2n-3)\cdots 3\cdot1\\ \end{align*}

We obtain \begin{align*} \binom{-\frac{1}{2}}{n}&=\frac{\left(-\frac{1}{2}\right)\left(-\frac{1}{2}-1\right) \cdots\left(-\frac{1}{2}-n+1\right)}{n!}\tag{2}\\ &=\frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right) \cdots\left(-\frac{2n-1}{2}\right)}{n!}\\ &=\frac{(-1)^n}{2^n}\cdot\frac{1\cdot 3\cdots \ \left(2n-1\right)}{n!}\tag{3}\\ &=\frac{(-1)^n}{2^n}\cdot\frac{(2n-1)!!}{n!}\tag{4}\\ &=\frac{(-1)^n}{2^n}\cdot\frac{(2n)!}{(2n)!!n!}\tag{5}\\ &=\frac{(-1)^n}{2^n}\cdot\frac{(2n)!}{2^nn!n!}\tag{6}\\ &=\frac{(-1)^n(2n)!}{2^{2n}n!n!}\tag{7}\\ &=\frac{(-1)^n(2n-1)!}{2^{2n-1}(n-1)!n!}\tag{8}\\ \end{align*}


  • In (2) we use the definition of the binomial coefficient with $\alpha=-\frac{1}{2}$

  • In (3) we factor out $-\frac{1}{2}$ from each of the $n$ factors giving $(-1)^n\frac{1}{2^n}$

  • In (4) we use double factorials $(2n-1)!!$

  • In (5) we apply $$(2n)!=(2n)!!(2n-1)!!$$

  • In (6) we use the formula from (5) and note that

$$(2n)!!=(2n)(2n-2)\cdots4\cdot 2=2^nn!$$

  • In (7) we collect terms.

  • In (8) we devide numerator and denominotor by $2n$.

Note there is a factor $2^{2n-1}$ in the denominator of $a_n$ which gives for $n=1$ \begin{align*} \binom{-\frac{1}{2}}{1}=-\frac{1}{2} \end{align*}


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