How to simplify hypergeometrics

*To*: mathgroup at smc.vnet.net*Subject*: [mg128379] How to simplify hypergeometrics*From*: "Dr. Wolfgang Hintze" <weh at snafu.de>*Date*: Thu, 11 Oct 2012 02:08:47 -0400 (EDT)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*Delivered-to*: l-mathgroup@wolfram.com*Delivered-to*: mathgroup-newout@smc.vnet.net*Delivered-to*: mathgroup-newsend@smc.vnet.net

Consider the probability function p[n_, a_, b_, k_] := Binomial[k - 1, a - 1]*(Binomial[n - k, b - a]/Binomial[n, b]) /; {0 <= a <= b, n >= a} In[68]:= p[n, a, b, k] Out[68]= (Binomial[-1 + k, -1 + a]*Binomial[-k + n, -a + b])/ Binomial[n, b] Now let's look for the zeroeth moment k^0 (for the higher ones the situation is similar) In[69]:= k0 = Sum[p[n, a, b, k], {k, 1, n}] Out[69]= ((-Binomial[-1, -a + b])*Binomial[n, -1 + a]* Hypergeometric2F1[1 - a + b, 1 + n, 2 - a + n, 1] + Binomial[0, -1 + a]*Binomial[-1 + n, -a + b]* HypergeometricPFQ[{1, 1, 1 - a + b - n}, {2 - a, 1 - n}, 1])/ Binomial[n, b] This should give 1, but it looks clumsy. Simplifying simply gives In[71]:= FullSimplify[k0] Out[71]= ComplexInfinity but also a qualified Simplify does not help, because in this case Mathematica makes slight cosmetic changes but the result is still far from being recognized as 1: In[73]:= FullSimplify[k0, {Element[{a, b, n}, Integers], 0 <= a <= b, n >= a}] Out[73]= ((-(-1)^(a + b))*Binomial[n, -1 + a]* Hypergeometric2F1[1 - a + b, 1 + n, 2 - a + n, 1] + Binomial[0, -1 + a]*Binomial[-1 + n, -a + b]* HypergeometricPFQ[{1, 1, 1 - a + b - n}, {2 - a, 1 - n}, 1])/ Binomial[n, b] Any help would be greatly appreciated. Best regards, Wolfgang