What is wrong with this calculation of $\binom{\frac{1}{2}}{k}$?

Question

The reason I ask this question is that I want to show that: \begin{equation*} \binom{\frac{1}{2}}{k} = -2\frac{1}{k}\binom{2k-2}{k-1}\left(-\frac{1}{4} \right)^k \end{equation*}

\begin{align*} \binom{\frac{1}{2}}{k} &= \frac{1}{k!}\left[ \frac{1}{2} \left(\frac{1}{2} - 1 \right) \left(\frac{1}{2} - 2 \right) \cdots \left(\frac{1}{2} - k + 1 \right)\right] \\ &= \frac{1}{k!}\frac{1}{2}\left[ \left(-\frac{1}{2}\right) \left(-\frac{3}{2}\right) \left(-\frac{5}{2}\right) \cdots \left(\frac{1-2k+2}{2}\right) \right] \\ &= \frac{1}{k!}\frac{1}{2}\left[ \left(-\frac{1}{2}\right) \left(-\frac{3}{2}\right) \left(-\frac{5}{2}\right) \cdots \left(\frac{1-2k+2}{2}\right) \right] \\ &= \frac{1}{k!}\frac{1}{2}\left[ \left(-\frac{1}{2}\right) \left(-\frac{3}{2}\right) \left(-\frac{5}{2}\right) \cdots \left(\frac{-2k+3}{2}\right) \right] \\ &= \frac{1}{k!}\frac{1}{2}\left[ \left(-\frac{1}{2}\right) \left(-\frac{3}{2}\right) \left(-\frac{5}{2}\right) \cdots \left(-\frac{2k - 3}{2}\right) \right] \\ &= \frac{1}{k!}\frac{1}{2}(-1)^{k - 1} \frac{1}{2^{k - 1}}\left(1 \times 3 \times 5 \times \cdots \times (2k - 3) \right) \\ &= \frac{1}{k!} \frac{1}{2} \left( -\frac{1}{2} \right)^{k - 1} \left(1 \times 3 \times 5 \times \cdots \times (2k - 3) \right) \\ &= \frac{1}{k!} \frac{1}{2} \left( -\frac{1}{2} \right)^{k - 1} \frac{1 \times 2 \times 3 \times 4 \times \cdots \times (2k - 3)}{2 \times 4 \times 6 \times \cdots \times (2k - 2)} \\ &= \frac{1}{k!} \frac{1}{2} \left( -\frac{1}{2} \right)^{k - 1} \frac{1 \times 2 \times 3 \times 4 \times \cdots \times (2k - 3)}{2 \times 4 \times 6 \times \cdots \times (2k - 2)} \\ &= \frac{1}{k!} \frac{1}{2} \left( -\frac{1}{2} \right)^{k - 1} \frac{1 \times 2 \times 3 \times 4 \times \cdots \times (2k - 3)}{2^{k - 1} (k - 1)!} \\ &= \frac{1}{k!} \frac{1}{2} \left( -\frac{1}{2} \right)^{k - 1} \left( -\frac{1}{2} \right)^{k - 1} (-1)^{k-1} \frac{1 \times 2 \times 3 \times 4 \times \cdots \times (2k - 3)}{(k - 1)!} \\ &= \frac{1}{k!} \frac{1}{2} \left( -\frac{1}{2} \right)^{2k - 2} (-1)^{k-1} \frac{1 \times 2 \times 3 \times 4 \times \cdots \times (2k - 3)}{(k - 1)!} \\ &= \frac{1}{k!} \frac{1}{2} \left( \frac{1}{4} \right)^{k - 1} (-1)^{k-1} \frac{1 \times 2 \times 3 \times 4 \times \cdots \times (2k - 3)}{(k - 1)!} \\ &= \frac{1}{k!} \frac{1}{2} \left( -\frac{1}{4} \right)^{k - 1} \frac{1 \times 2 \times 3 \times 4 \times \cdots \times (2k - 3)}{(k - 1)!} \\ &= \frac{1}{2} \left( -\frac{1}{4} \right)^{k - 1} \frac{1 \times 2 \times 3 \times 4 \times \cdots \times (2k - 3)}{k!(k - 1)!} \\ &= \frac{1}{2} \left( -\frac{1}{4} \right)^{k - 1} \frac{1 \times 2 \times 3 \times 4 \times \cdots \times (2k - 3)}{k(k-1)!(k - 1)!} \\ &= \frac{1}{2} \left( -\frac{1}{4} \right)^{k - 1} \frac{(2k - 3)!}{k(k-1)!(k - 1)!} \\ \end{align*}

However, the last step in the proof above has $(2k-3)!$ in the numerator instead of $(2k-2)!$ in the numerator.

Answer 1

A definition of binomial coefficient that suits your computation follows naturally from $$\binom{n}{k} = \frac{n!}{k! (n-k)!} = \frac{n(n-1)\ldots(n-k+1)}{k!} = \prod_{m=1}^k \frac{n-m+1}{m},$$ which works when the lower index is a nonnegative integer (if $k = 0$, then the product is empty and equals 1).

Consequently, $$\begin{align*} \binom{1/2}{k} &= \prod_{m=1}^k \frac{1/2-m+1}{m} = \frac{1}{2} \prod_{m=2}^k \frac{3-2m}{2m} \\ &= \frac{1}{2} \prod_{m=1}^{k-1} \frac{-(2m-1)(2m)}{4m(m+1)} \\ &= \tfrac{1}{2}\left(-\tfrac{1}{4}\right)^{k-1} \frac{(2(k-1))!}{(k-1)! k!} \\ &= -\frac{2}{k}(-4)^{-k} \binom{2(k-1)}{k-1}. \end{align*}$$

Answer 2

Using \begin{align} \binom{a}{k} = \frac{(-1)^{k}}{k!} \cdot \frac{\Gamma(k-a)}{\Gamma(-a)} \end{align} then \begin{align} \binom{\frac{1}{2}}{k} = \frac{(-1)^{k+1}}{\sqrt{\pi} k!} \cdot \frac{\Gamma\left(k + \frac{1}{2}\right)}{2k-1}. \end{align} Using \begin{align} \sqrt{2 \pi} \Gamma(2x+1) = 2^{2x+1/2} \Gamma(x+1) \Gamma\left(x + \frac{1}{2}\right) \end{align} then \begin{align} \binom{\frac{1}{2}}{k} = \frac{(-1)^{k+1}}{4^{k}} \binom{2k}{k}. \end{align}

Answer 3

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{{1/2 \choose k} = -\,{2 \over k}\,{2k - 2 \choose k - 1}\pars{-\,{1 \over 4}}^{k}:\ {\large ?}}$.

\begin{align} &\mbox{Whith}\ \verts{x} < {1 \over 4}\,,\quad \color{#c00000}{\sum_{k\ =\ 1}^{\infty}x^{k - 1}{2k - 2 \choose k - 1}} =\sum_{k\ =\ 1}^{\infty}x^{k - 1} \oint_{\verts{z}\ =\ 1^{-}}{\pars{1 + z}^{2k - 2} \over z^{k}} \,{\dd z \over 2\pi\ic} \\[5mm]&=\oint_{\verts{z}\ =\ 1^{-}}{1 \over z}\sum_{k\ =\ 1}^{\infty} \bracks{\pars{1 + z}^{2}x \over z}^{k - 1}\,{\dd z \over 2\pi\ic} =\oint_{\verts{z}\ =\ 1^{-}}{1 \over z}\, {1 \over 1 - \pars{1 + z}^{2}x/z}\,{\dd z \over 2\pi\ic} \\[5mm]&=-\oint_{\verts{z}\ =\ 1^{-}}{1 \over x\pars{1 + z}^{2} - z} \,{\dd z \over 2\pi\ic} =-\,{1 \over x}\oint_{\verts{z}\ =\ 1^{-}} {1 \over \pars{z - r_{+}}\pars{z - r_{-}}}\,{\dd z \over 2\pi\ic} \end{align}

where $\ds{r_{\pm} \equiv {1 \pm \root{1 - 4x} - 2x \over 2x}}$

Then, \begin{align} &\color{#c00000}{\sum_{k\ =\ 1}^{\infty}x^{k - 1}{2k - 2 \choose k - 1}} =-\,{1 \over x}\,{1 \over r_{-}\ -\ r_{+}} =-\,{1 \over x}\,{1 \over -\root{1 - 4x}/x} = {1 \over \root{1 - 4x}} \end{align}

That yields $$ \color{#c00000}{\sum_{k\ =\ 1}^{\infty}{x^{k} \over k}\,{2k - 2 \choose k - 1}} = \int_{0}^{x}{\dd t \over \root{1 - 4t}} = \half\pars{1 - \root{1 - 4x}} $$

We'll make the replacement $\ds{x\ \mapsto\ -\,{1 \over 4}\,x}$: \begin{align} &\color{#c00000}{-2\sum_{k\ =\ 1}^{\infty}{\pars{-1/4}^{k} \over k}\, {2k - 2 \choose k - 1}x^{k}} =\root{1 + x} - 1\quad\mbox{where}\quad\verts{x} < 1 \end{align}

Expanding the right hand side: $$ \color{#c00000}{\sum_{k\ =\ 1}^{\infty}{-2 \over k}\,\pars{-\,{1 \over 4}}^{k} {2k - 2 \choose k - 1}x^{k}} =\sum_{k = 1}^{\infty}{1/2 \choose k}x^{k} $$

Then $$ \color{#66f}{\large% {1/2 \choose k} =-2\,{1 \over k}\,{2k - 2 \choose k - 1}\pars{-\,{1 \over 4}}^{k}} $$