# Simplify doesn't simplify HypergeometricPFQ with exact arguments

Consider a series: $$\sum_{t=0}^\infty \frac{8^{-11-2t}(22+4t)!}{t!(11+t)!(11+2t)!(32+t)}$$

Sum[(8^(-11 - 2 t) (22 + 4 t)!)/(t! (11 + t)! (11 + 2 t)!) 1/(32 + t),{t,0,Infinity}]


I enter it into Mathematica, and get the following: $$\frac{26880307030942914706481517908268600094665992\sqrt2}{2617883526123366341980485070559163273300330975\pi}\tag1$$

26880307030942914706481517908268600094665992 Sqrt/(2617883526123366341980485070559163273300330975 Pi)


So far so good. Now I change $32$ to q and get the new answer: $$\frac{88179{\;}_3 F_2\left(\frac{23}4,\frac{25}4,q;12,q+1;1\right)}{1073741824q}\tag2$$

88179 HypergeometricPFQ[{23/4, 25/4, q}, {12, 1 + q}, 1]/(1073741824 q)


OK, not bad. Let's now do %/.q->32. But... it remains in almost unchanged form of $$\frac{88179{\;}_3 F_2\left(\frac{23}4,\frac{25}4,32;12,33;1\right)}{34359738368}\tag3$$ I now try doing Simplify, FullSimplify, but it doesn't change. I tried checking $(1)$ and $(2)$ for equality with ==, and still got unevaluated expression. Using Reduce over it gave me True with strange problem of

Reduce::ztest1: Unable to decide whether numeric quantity /* difference of LHS and RHS multiplied by denominator */ is equal to zero. Assuming it is. >>

Still, computing difference between ${}_3 F_2\left(\frac{23}4,\frac{25}4,32;12,33;1\right)$ and $\frac{923600316834709429658044469155615723122343517467181056\sqrt2}{230842351450032320669497193036836458276349885044525\pi}$ with N[...,30] gives me almost zero ($10^{-74}$), so this function indeed must be simplifiable.

So, how do I convince Mathematica to simplify such hypergeometric function expressions?

• Just an observation, FullSimplify[(sum2 /. q -> 32) - sum1] gives 0 without warnings. Dec 7, 2013 at 9:15
• Oh, indeed, using FullSimplify instead of Reduce gives True without warnings. Still, it doesn't reveal $(1)$ without need to put it into expression. Dec 7, 2013 at 9:17
• @Nasser added some code Dec 7, 2013 at 9:56

expr2 = Sum[(8^(-11 - 2 t) (22 + 4 t)!)/(t! (11 + t)! (11 +2 t)!) 1/(q + t), {t, 0, Infinity}] FunctionExpand[%] • Wow, didn't know of FunctionExpand. This is great. Thanks. Dec 7, 2013 at 10:02