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How to find a simplified expression for this specific binomial coefficient?

$$\binom{\frac{1}{2}}{n}$$

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closed as off-topic by heropup, JMP, Kamil Jarosz, Claude Leibovici, 3SAT Feb 13 '16 at 11:05

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  • $\begingroup$ I've edited your question to format it. Please make sure it says what you intended. $\endgroup$ – user296602 Feb 13 '16 at 2:56
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    $\begingroup$ Isn't this what you're talking about? $\endgroup$ – Jared Feb 13 '16 at 3:02
  • $\begingroup$ Related math.stackexchange.com/questions/396889/… $\endgroup$ – cgiovanardi Feb 13 '16 at 10:44
  • $\begingroup$ Why all the downvotes on the answers? $\endgroup$ – robjohn Feb 17 '16 at 22:31
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I do not know if the following expression is that you are looking for

$$\frac12\frac{\sqrt\pi}{\Gamma(1+n)\Gamma\left(\frac32-n\right)}$$

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We have

\begin{align} {{1/2}\choose n}&=\frac{(1/2)(1/2-1)(1/2-2)\cdots(1/2-n+1)}{n!}\\ \\ &=\frac{(1/2)(-1/2)(-3/2)\cdots((3-2n)/2)}{n!}\\ \\ &=\frac{(1)(-1)(-3)\cdots(3-2n)}{2^{n}n!}\\ \\ &=(-1)^{n-1}\frac{(1)(3)(5)\cdots(2n-3)}{2^nn!}\\ \\ &=(-1)^{n-1}\frac{(2n-3)!}{2^nn!(2)(4)(6)\cdots(2n-2)}\\ \\ &=(-1)^{n-1}\frac{(2n-3)!}{2^n2^{n-1}n!(n-1)!} \end{align}

I feel like I almost certainly miscounted a few things there but the point is that you can get a closed form with "simple" factors.

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As $\binom{n}{k}=\frac{n(n-1)(n-2)...(n-k+1)}{k!}$, we can write:

$$\binom{\frac{1}{2}}{n}=\frac{(1/2)(-1/2)(-3/2)...(3/2 - n)}{n!}=\frac{(-1)^{n-1}(1)(3)(5)...(2n-3)}{n!2^n}$$

$$=\frac{(-1)^{n-1}(1)(2)(3)(4)(5)...(2n-3)(2n-2)}{n!2^n(2)(4)(6)...(2n-2)}=\frac{(-1)^{n-1}(2n-2)!}{2^{2n-1}n!(n-1)!}=\frac{(-1)^{n-1}(2n)!}{2^{2n}n!n!(2n-1)}$$

$$=\frac{(-1)^{n-1}\binom{2n}{n}}{2^{2n}(2n-1)}$$

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$$ \begin{align} \frac{\frac12\cdot\left(-\frac12\right)\cdots\left(\frac{3-2n}2\right)}{n!} &=\frac{(-1)^{n-1}}{2^nn!}(2n-3)(2n-5)\cdots1\\ &=\frac{(-1)^{n-1}}{2^nn!}\frac{(2n-2)!}{2^{n-1}(n-1)!}\\ &=\frac{(-1)^{n-1}}{2^{2n-1}n}\binom{2n-2}{n-1} \end{align} $$

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One useful form for that binomial coefficient is obtained by expanding it out and observing some patterns. Since $\binom{1/2}{n}=\frac{\frac{1}{2}}{n}\cdot\frac{\frac{1}{2}-1}{n-1}\cdots\frac{\frac{1}{2}-n+1}{1}=\frac{1}{2n}\cdot\frac{-1}{2(n-1)}\cdot\frac{-3}{2(n-2)}\cdots\frac{3-2n}{2}$, since there are $n-1$ factors of $-1$, and by reversing the order of the denominators, we get $\frac{(-1)^{n-1}}{2n}\cdot\frac{1}{2}\frac{3}{4}\frac{5}{6}\cdots\frac{2n-3}{2n-2}$.

Or, the factors of $2$ may be accumulated to write $\frac{(-1)^{n-1}}{2^nn!}\cdot 3\cdot5\cdot7\cdots(2n-3)$.

This product of odd numbers can be derived from dividing $(2n-3)!$ by $2^{n-2}(n-2)!$.

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