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I think the following identity is true, $$ \frac{4 (4 s+9)}{3 \Gamma \left(s+\frac{5}{2}\right) \Gamma \left(s+\frac{7}{2}\right)}-\frac{16 (s+2)}{3 \Gamma (s+3)^2}=\frac{\, _3F_2\left(2,s+\frac{5}{2},s+\frac{7}{2};s+4,s+5;1\right)}{\Gamma (s+4) \Gamma (s+5)} $$ But I haven't found anything useful in the literature.

This identity is inspired by this question.

NonalcoholicBeer
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    Since $$,3F_2(a_1,a_2,a_3;b_1,b_2;1)=\sum{n\geq0}\frac{(a_1)n(a_2)_n(a_3)_n}{(b_1)_n(b_2)_n n!}$$ And since $(x)_n=\frac{\Gamma(x+n)}{\Gamma(x)}$ you can factor: $$,_3F_2(a_1,a_2,a_3;b_1,b_2;1)=\frac{\Gamma(b_1)\Gamma(b_2)}{\Gamma(a_1)\Gamma(a_2)\Gamma(a_3)}\sum{n\geq0}\frac{\Gamma(a_1+n)\Gamma(a_2+n)\Gamma(a_3+n)}{\Gamma(b_1+n)\Gamma(b_2+n) n!}$$ which will help you simplify things – clathratus Jan 15 '19 at 01:08
  • This is only valid if s+5/2 is a nonpositive integer, right? – NonalcoholicBeer Jan 15 '19 at 14:43
  • I think it's valid for $$\text{Re},s>-\frac52$$ – clathratus Jan 15 '19 at 15:25
  • Okay. For anyone who's interested in a proof, see Doron Zeilberger's book A=B, example 7.3.3. – NonalcoholicBeer Jan 18 '19 at 17:19

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