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Prove the (numerically-evident) proposition that \begin{equation} \Sigma_{i=0}^\infty f(i) = 1, \end{equation} where \begin{equation} f(i)= 2^{-4 i-6} q(i) \frac{\Gamma(3 i+\frac{5}{2}) \Gamma(5 i+2)}{3 \Gamma(i+1) \Gamma(2 i+3) \Gamma(5 i+\frac{13}{2})} \end{equation} and \begin{equation} q(i) = 185000 i^5 +779750 i^4 +1289125 i^3 +1042015 i^2 +410694 i+63000= \end{equation} \begin{equation} i (5 i (25 i (2 i (740 i+3119)+10313)+208403)+410694)+63000. \end{equation} This is the special case ($\alpha=0$) of eqs. (4)-(6) in arXiv:1301.6617, and eqs. (2)-(3) in arXiv:1303.1125.

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Is this a cry for help or an invitation to a contest? – Rhett Butler May 9 '13 at 15:23
Have you tried the Gosper-Zeilberger algorithm for summing hypergeometric series? – Douglas Zare May 9 '13 at 17:57
can you give some background how this identity to unity arises? does that suggest a way to a proof? – Carlo Beenakker May 9 '13 at 19:57
Then too, some numerically evident propositions are false... For example, $$ \sum_{n=1}^\infty \frac{(3n-2)!}{(2n)!} \frac{(2n+99)!}{(3n+99)!} $$ is not quite equal to what numerical computation suggests. See . – Noam D. Elkies May 9 '13 at 22:06
Denominator 16 ... maybe it is somehow related to these:… – Gerald Edgar May 12 '13 at 13:10

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