That peculiar series I've been hacking boils down to (c43) (hyper_f[4,3]([1,a,b,c],[1-a,1-b,1-c]),%%=multthru(hypersimp(%%))) Time= 50 msec. (d43) hyper_f ([1, a, b, c], [1 - a, 1 - b, 1 - c]) = 4, 3 gamma(1 - a) gamma(1 - b) gamma(1 - c) gamma(- c - b - a + 1) 1 ------------------------------------------------------------- + - 2 gamma(- b - a + 1) gamma(- c - a + 1) gamma(- c - b + 1) 2 which, as you can see, I've now automated, in spite of the absence of a contiguous system. Thus, you typically get a gamma product + a rational term + a canonical 4F3 with parameters in the form [1,x,y,z; 1-x,1-y,2-z] : (c54) (hyper_f[4,3]([1,a,b,c],[-a,-b,-c]),%%=hypersimp(%%)) Time= 200 msec. (d54) hyper_f ([1, a, b, c], [- a, - b, - c]) = 4, 3 (a b c (c + b + a + 2) (2 c + 2 b + 2 a + 3) gamma(- a) gamma(- b) gamma(- c - b - a - 2) gamma(- c) (-------------------------------------------------------- + 1) gamma(- b - a - 1) gamma(- c - a - 1) gamma(- c - b - 1) /((b + a + 1) (c + a + 1)) + (b + 1) (c + 1) 2 2 2 2 2 (4 b c + 2 a c + 2 c + 4 b c + 6 a b c + 8 b c + 2 a c + 5 a c + 3 c 2 2 2 2 + 2 a b + 2 b + 2 a b + 5 a b + 3 b + a + 2 a + 1) /((b + a + 1) (c + a + 1)) - 2 hyper_f ([1, c + 1, b + 1, a + 1], 4, 3 [- c, - b, 1 - a], 1) b c (2 c + 2 b + 2 a + 3)) /((2 b + 1) (c + b + 1) (2 c + 1)) This isn't much use unless you luck out with a zero coefficient on the 4F3: (c50) (sum(i^2*(i-a-1)!*(i-b-1)!*(i-c-1)!/((i+a)!*(i+b)!*(i+c)!),i,1,inf), resimplify(%%=make1hyper(%%))=factcomb(factor(maxfactorial(makefact(hypersimp(%%)))))) Time= 34879 msec. inf ==== 2 \ i (i - a - 1)! (i - b - 1)! (i - c - 1)! (d50) > ----------------------------------------- = / (i + a)! (i + b)! (i + c)! ==== i = 1 3 %pi hyper_f ([2, 2, 1 - a, 1 - b, 1 - c], [1, a + 2, b + 2, c + 2], 1) 5, 4 /((a - 1)! (a + 1)! sin(%pi a) (b - 1)! (b + 1)! sin(%pi b) (c - 1)! (c + 1)! 3 sin(%pi c)) = %pi (c + b + a - 1)!/(2 (a - 1)! sin(%pi a) (b - 1)! sin(%pi b) (b + a)! (c - 1)! sin(%pi c) (c + a)! (c + b)!) Note that this took half a minute and extensive post-simplification, and needs a bit more: (c51) ((-a)!*(-b)!*(-c)!,(1=%%)=%%,factcomb(negfactorial(%/%%)*%%)) Time= 60 msec. inf ==== 2 \ i (i - a - 1)! (i - b - 1)! (i - c - 1)! (d51) > ----------------------------------------- = / (i + a)! (i + b)! (i + c)! ==== i = 1 hyper_f ([2, 2, 1 - a, 1 - b, 1 - c], [1, a + 2, b + 2, c + 2], 1) (- a)! 5, 4 (- b)! (- c)!/((a + 1)! (b + 1)! (c + 1)!) = (- a)! (- b)! (- c)! (c + b + a - 1)! -------------------------------------, 2 (b + a)! (c + a)! (c + b)! a monomial! This is strange because the base case is a binomial, but due to the lack of "left termination", the three term contiguity relations are inhomogeneous, which adds in rational terms. Getting this calculation to even fit in memory, let alone in 34 seconds, was a bear to program. I need another 4F3[1] identity to recoup the effort! Apropos surprisingly messy results, one can easily sum certain hybrid trig/hypergeometric series by, e.g., inf inf ==== ==== %i n x \ sin(n x) \ %e (d60) > ---------------- = Im( > ----------------) / binomial(2 n, n) / binomial(2 n, n) ==== ==== n = 0 n = 0 but the answer, which is real, will look complex. Macsyma's IMAGPART can nearly always eliminate the "%i"s, but leaves a big mess. Extensive chiropractic got it down to inf ==== 2 \ sin(n x) 2 cos (x) - 76 cos(x) + 47 (d158) > ---------------- = sqrt(2) sqrt(-------------------------- + 1) / binomial(2 n, n) 3/2 ==== (17 - 8 cos(x)) n = 0 sqrt(4 cos(x) + sqrt(17 - 8 cos(x)) - 1) 3/4 asin(----------------------------------------)/(17 - 8 cos(x)) 2 sqrt(2) 2 2 cos (x) - 76 cos(x) + 47 + sqrt(2) sqrt(1 - --------------------------) 3/2 (17 - 8 cos(x)) sqrt(- 4 cos(x) + sqrt(17 - 8 cos(x)) + 1) 3/4 asinh(------------------------------------------)/(17 - 8 cos(x)) 2 sqrt(2) 4 sin(x) + ------------- 17 - 8 cos(x) (Seventeen? Forty seven??) Symbolically testing: c159) taylor(%,x,0,5) inf ==== 3 \ n 3 ( > ----------------) x inf / binomial(2 n, n) ==== ==== \ n n = 0 (d159)/T/ ( > ----------------) x - --------------------------- / binomial(2 n, n) 6 ==== n = 0 inf ==== 5 \ n 5 ( > ----------------) x / binomial(2 n, n) ==== n = 0 + --------------------------- + . . . = 120 3 (2 %pi sqrt(3) + 18) x (9 sqrt(6) sqrt(3) sqrt(2) + 37 %pi sqrt(3) + 351) x ---------------------- - ----------------------------------------------------- 27 729 5 (445 sqrt(6) sqrt(3) sqrt(2) + 1407 %pi sqrt(3) + 12639) x + ----------------------------------------------------------- + . . . 43740 Doing the series: (c162) block([algebraic:true],expand(ratsimp(hypersimp(%)))) 5 5 3 3 469 sqrt(3) %pi x 7 x 37 sqrt(3) %pi x 5 x 2 sqrt(3) %pi x (d160) ------------------ + ---- - ----------------- - ---- + --------------- 14580 20 729 9 27 5 5 3 3 2 x 469 sqrt(3) %pi x 7 x 37 sqrt(3) %pi x 5 x 2 sqrt(3) %pi x + --- = ------------------ + ---- - ----------------- - ---- + --------------- 3 14580 20 729 9 27 2 x + ---, so it has my vote. 3 With x = pi/2, we get another one of those screwy hypergeometrics with a 1 upstairs to cancel the familiar n! out of the denominator: 3 3 5 1 hyper_f ([1, 1, -], [-, -], - --) 3, 2 2 4 4 16 (d171) ------------------------------------ = 2 47 sqrt(sqrt(17) + 1) sqrt(2) sqrt(sqrt(17) - --) asinh(------------------) 17 2 sqrt(2) ----------------------------------------------------- 17 47 sqrt(sqrt(17) - 1) sqrt(2) sqrt(sqrt(17) + --) asin(------------------) 17 2 sqrt(2) 4 + ---------------------------------------------------- + -- 17 17 (c172) dfloat(%) (d172) 0.45369611887698d0 = 0.45369611887698d0 I was anxious to try Mma's big hammer simplifiers on (d158), but couldn't figure out how to get past the first step of syntactically eliminating the imaginary units. Can some Mma wizard explain how to get Re[ArcSin[2]], e.g., (c61) realpart(asin(2)) %pi (d61) --- 2 Maple also does this case, but my V.4 can't seem to do the mess fom the sum, nor even compute the mess in the first place. Meanwhile, I remain convinced that simplifications such as (d158) are an open AI problem. Computer algebra and AI parted ways with the Risch and Berlekamp algorithms. Maybe it's time they got back together.