[math-fun] Everybody needs a conversation stopper.

Sum[ArcTanh[√3Tan[t/(-2)^n]], {n,∞}] == -ArcTanh[Tan[t]/√3] Actually, this clearly must telescope, with a presumably easy inductive verification. It might even be worthwhile for Mathematica to check for summands and prodands of the form f(index+freevariable). --rwg