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)!}

closed as off-topic by Daniel Lichtblau, corey979, Coolwater, JimB, MarcoB Aug 6 '18 at 4:16

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question cannot be answered without additional information. Questions on problems in code must describe the specific problem and include valid code to reproduce it. Any data used for programming examples should be embedded in the question or code to generate the (fake) data must be included." – Daniel Lichtblau, Coolwater, JimB
If this question can be reworded to fit the rules in the help center, please edit the question.