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Timeline for Proving a binomial sum identity

Current License: CC BY-SA 4.0

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Jul 27, 2021 at 18:57 vote accept T. Amdeberhan
Jul 27, 2021 at 17:24 history edited T. Amdeberhan CC BY-SA 4.0
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Jul 27, 2021 at 13:26 comment added T. Amdeberhan That's also an interesting proposition.
Jul 27, 2021 at 5:43 comment added Alexander Burstein @T.Amdeberhan I wonder if there is a nice probabilistic interpretation of $$\sum_{n=0}^{\infty}{\frac{\binom{2n}{n}\binom{2x}{x}x}{2^{2(n+x)}(n+x)}}=1.$$
Jul 26, 2021 at 21:04 comment added T. Amdeberhan @AlexanderBurstein: thanks, this is a neater form.
Jul 26, 2021 at 13:28 comment added T. Amdeberhan @Nemo: One of my tags is "soft question". Nonetheless, look at the rich and pedagogical responses.
Jul 26, 2021 at 13:23 comment added Ira Gessel The sum is $x^{-1}(\,{}_2F_1(1/2,𝑥;1+𝑥\mid 1)−1)$, which can be evaluated by Gauss's theorem, one of the standard hypergeometric summation theorems. Max Alekseyev's proof is essentially a special case of the usual proof of Gauss's theorem.
Jul 26, 2021 at 10:41 review Close votes
Jul 29, 2021 at 7:35
Jul 26, 2021 at 10:28 comment added Nemo This is elementary textbook material. See exercise 8 to chapter 12 of Whittaker and Watson.
Jul 26, 2021 at 9:37 answer added Alapan Das timeline score: 4
Jul 26, 2021 at 8:08 history edited Martin Sleziak
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Jul 26, 2021 at 3:44 comment added Alexander Burstein This can be rewritten as $$\sum_{n=0}^{\infty}\frac{\binom{2n+2}{n+1}}{2^{2n+2}\,(n+x+1)}=\frac{2^{2x}}{x\,\binom{2x}x}-\frac{1}{x},$$ or equivalently, $$\sum_{n=0}^{\infty}\frac{\binom{2n}{n}}{2^{2n}\,(n+x)}=\frac{2^{2x}}{x\,\binom{2x}x}.$$
Jul 26, 2021 at 0:42 history became hot network question
Jul 25, 2021 at 22:52 answer added Max Alekseyev timeline score: 9
Jul 25, 2021 at 17:41 answer added user127776 timeline score: 6
Jul 25, 2021 at 17:24 answer added Carlo Beenakker timeline score: 3
Jul 25, 2021 at 16:37 history asked T. Amdeberhan CC BY-SA 4.0