Is the following evaluation of the sum correct?
$$\sum_{n=0}^k \frac{(\frac{1}{2})_n}{n!} \frac{(\frac{1}{2})_{k-n}}{(k-n)!} =1.$$
I don't see how this is immediate from the definitions.
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You can easily check that for $k=2$:
$$\frac 1{0!}\frac {\left(\frac12\right)\left(-\frac12\right)}{2!}+\frac {\frac12}{1!}\frac{\frac12}{1!}+\frac {\left(\frac12\right)\left(-\frac12\right)}{2!}\frac 1{0!}=0\ne1$$
In fact: $$\sum_{n=0}^k \frac{(\frac{1}{2})_n}{n!} \frac{(\frac{1}{2})_{k-n}}{(k-n)!} =\sum_{n=0}^k\binom {\frac12}n\binom{\frac12}{k-n} = \binom {\frac12+\frac12}k=\binom1k=\begin{cases}1 &\text{for }k=0,1\\0 &\text{otherwise}\end{cases}$$
The equality used is the Chu-Vandermonde Identity.
player3236
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You can see that Wikipedia has a wrong Pochhammer Symbol definition. You can compare, for instance, with MathWorld and DLMF. – Felix Marin Sep 16 '20 at 20:58
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Says the person who didn't even read through the second paragraph of the MathWorld definition. – player3236 Sep 17 '20 at 01:45
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the case k=2 equals to 1, check wolfram alpha – User1825 Sep 17 '20 at 03:43
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{equation} \mbox{Note that}\quad \pars{a}_{n} = a^{\overline{n}} = {\Gamma\pars{a + n} \over \Gamma\pars{a}} = {\pars{a + n - 1}! \over \pars{a - 1}!}\label{1}\tag{1} \end{equation} In (\ref{1}), I'm using the DLMF, MathWorld and Wolfram-${\tt Mathematica}$ definitions.
Wikipedia, I guess, has a wrong definition $\require{cancel}\ds{\cancel{\pars{x}_{n} = x^{\underline{n}} = x\pars{x - 1}\ldots\pars{x - n + 1}}}$.
With (\ref{1}): \begin{align} &\bbox[5px,#ffd]{\sum_{n = 0}^{k}{\pars{1/2}_{n} \over n!}\, {\pars{1/2}_{k - n} \over \pars{k - n}!}} = \sum_{n = 0}^{\color{red}{\infty}}{\pars{n - 1/2}!\,/\pars{-1/2}! \over n!}\, {\pars{k - n - 1/2}!\,/\pars{-1/2}! \over \color{red}{\pars{k - n}!}} \\[5mm] = &\ \sum_{n = 0}^{\infty}{n - 1/2 \choose n}{k - n - 1/2 \choose k - n} = \sum_{n = 0}^{\infty} \bracks{{-1/2 \choose n}\pars{-1}^{n}} \bracks{{-1/2 \choose k - n}\pars{-1}^{k - n}} \\[5mm] = &\ \pars{-1}^{k}\sum_{n = 0}^{\infty} {-1/2 \choose n}\bracks{z^{k - n}}\pars{1 + z}^{-1/2} \\[5mm] = &\ \pars{-1}^{k}\bracks{z^{k}}\pars{1 + z}^{-1/2} \sum_{n = 0}^{\infty}{-1/2 \choose n}z^{n} = \pars{-1}^{k}\bracks{z^{k}}\pars{1 + z}^{-1/2} \pars{1 + z}^{-1/2} \\[5mm] = &\ \pars{-1}^{k}\bracks{z^{k}}\pars{1 + z}^{-1} = \pars{-1}^{k}\bracks{\pars{-1}^{k}} = \bbx{\large 1} \\ & \end{align}
Felix Marin
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in the first line, how did you get the upper limit to be at infinity? – User1825 Sep 17 '20 at 04:57
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@User1825 $\left(z - n\right)! \to \infty$ whenever $z \to k$ for $k = n + 1, n + 2, n + 3, \ldots$. – Felix Marin Sep 17 '20 at 05:58