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It is possible transform the following sum as n tend to Infinity $$\sum _{k=0}^{\infty } \frac{2^{2 (k+1)-2} (-1)^{k-n+2} \Gamma \left(k+\frac{5}{2}\right) \Gamma (k+n+2) \, _2F_3\left(\frac{k}{2}+\frac{5}{4},\frac{k}{2}+\frac{7}{4};\frac{3}{2},\frac{k}{2}+\frac{3}{2},\frac{k}{2}+2;-\frac{1}{16}\right)}{\sqrt{\pi}\; \Gamma (2 (k+1)) \Gamma (k+3) \Gamma (-k+n+1)}=\sin \left(\frac{1}{2}\right)$$ Using symbolic Wilf-Zeilberger , I try to get A G[k,n] from F[k,n] and acceleration the series Thanks

((-1)^(2+k-n) 2^(-2+2 (1+k)) Gamma[5/2+k] Gamma[2+k+n] HypergeometricPFQ[{5/4+k/2,7/4+k/2},{3/2,3/2+k/2,2+k/2},-(1/16)])/(Sqrt[\[Pi]] Gamma[2 (1+k)] Gamma[3+k] Gamma[1-k+n])


\sum _{k=0}^n \frac{x^k \left(\sqrt{\pi } (k-n-1)! (k+n+2)!\right)}{2 k! \left(k+\frac{1}{2}\right)! (-n-1)! (n+2)!}
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closed as off-topic by Daniel Lichtblau, corey979, Coolwater, JimB, MarcoB Aug 6 '18 at 4:16

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