Also, by http://dspace.mit.edu/bitstream/handle/1721.1/6088/AIM-304.pdf p25: HarmonicNumber[p]==-Sum[(3*k + p)* Binomial[k - p - 1, k]/k/(k + p)/Binomial[2*k, k], {k, ∞}] (Terminates after (integer) p), which Mma converts to two useless 3F2s unless integer 2p. --rwg Ouch, there are way more typos in AIM 304 than I thought. Even "some" instead of "sum". I wonder if MIT lost the errata. Repairing one egregium, p66: Sum[1/Fibonacci[n], {n, ∞}] == Sum[(I^((-1 + k)*k)* ((-1)^k*Fibonacci[2 + 2*k] + Fibonacci[3 + 4*k]))/ (Fibonacci[1 + 2*k]*Fibonacci[2 + 2*k]* Product[LucasL[1 + 2*j], {j, k}]),{k, 0, ∞}] For some reason, this (rapidly convergent) rhs completely stymies Mma's numerics. It doesn't even yell False if you change LucasL[2j+1] to LucasL[2j-1]. There shortly follows an acceleration of Sum 1/(x^n+cy^n) which divides by zero if there is a nonnegative integer j s.t. x^j=y^(j+1) or y^j=x^(j+1). PUZZLE 1 (from back in my smarter days): rewrite Sum(1/(x^n-1)) (= Sum[DivisorSigma[0,n]/x^n]) so as to avoid the 1/0. SPOILER Sum[1/(x^n - 1), {n, ∞}] == Sum[1/(x^(n/4) - (-1)^n) + (-1)^n/(x^(n/4) - 1), {n, ∞}]/2 PUZZLE 2: Prove it. --rwg