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[2]/(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?
FullSimplify[(sum2 /. q -> 32) - sum1]
gives0
without warnings. $\endgroup$FullSimplify
instead ofReduce
givesTrue
without warnings. Still, it doesn't reveal $(1)$ without need to put it into expression. $\endgroup$