in http://gosper.org/fst.pdf, namely Sum[((-1)^n* Cos[Sqrt[(n + 1/2)^2 + a^2]*Pi])/((n + 1/2)*(n - a + 1/2)*(n + a + 1/2)), {n, 0, Infinity}] == (Pi*(Sin[a*Pi]^2 - 2*Sin[(a*Pi)/Sqrt[2]]^2))/(2*a^2*Cos[a*Pi]) gives oo=oo for a = 1/2+ (integer k), but gives (6*(-1)^k*Cos[((1 + 2*k)*Pi)/Sqrt[2]])/(1 + 2*k)^3 + ((-1)^k*Sqrt[2]*Pi*Sin[((1 + 2*k)*Pi)/Sqrt[2]])/(1 + 2*k)^2 if you skip the n=k term. --rwg With help from Corey and Julian, and buffoonery from Mathematica.
Consider the standard binary tree with infinitely many levels.
Suppose each edge is colored green with probability = p.
What is the probability f(p) that there exists an infinite green path starting at the root?
--Dan
----------------------------------------------------------------------- f(p) = (2p - 1)/p^2 for p > 1/2, f(p) = 0 for p <= 1/2. -- Gene Yow, slope 8 at p=1/2+ ! Maybe building a tree is a good way to test coins for fairness. --rwg