Is there an easy way to show that the following expression is non-zero only for $l=1$?
$$\sum_{k = 0}^{l} \binom{l}{k} \binom{\frac{1}{2}\left(l + k - 1\right)}{l} \frac{1 + (-1)^{k+1}}{k+2}$$
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How do you define a binomial coefficient at non-integral parameters (e.g. the middle one when $k=l$)? – Parcly Taxel Oct 18 '18 at 12:50
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@ParclyTaxel $\binom{x}{l}=\frac{x(x-1)\cdots(x-l+1)}{l!}$. It's commonly used for binomial Taylor expansion, for instance. – egreg Oct 18 '18 at 12:54
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Right, only the bottom parameter has to be integral for a finite expression. – Parcly Taxel Oct 18 '18 at 12:55
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But what does it mean for a binomial coefficient? if the upper number is smaller than the lower one, as it can happen in this sum? Select 3 balls out of 2? – Luke Oct 18 '18 at 13:14
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@Luke: By the expansion egreg quoted you would get $\binom 23=\frac{2\cdot 1\cdot 0}{3!}=0$ ways to select 3 balls out of 2. – hmakholm left over Monica Oct 18 '18 at 13:24
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(Don't ask me what sense it makes that there is $1$ way to select $4$ balls out of $-1$, though ...) – hmakholm left over Monica Oct 18 '18 at 13:34
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For negative numbers, the analytic extension of the Gamma function is used. Well, maybe I should rephrase the question: is there any way of showing this? – John Oct 18 '18 at 16:14