In[504]:= Simplify[Product[(1 + 1/(z - n)) (1 + 1/(z + n)), {n, ∞]}] (1 + 1/z)] == Limit[E^E^E^(x - E^(-E^E^x - E^x - x)) - E^E^E^x, x -> ∞] Out[504]= True I'm surprised lhs didn't demand noninteger z. —rwg From: Dan Asimov <dasimov@earthlink.net> Date: Sun, Oct 21, 2018, 22:51 Subject: [math-fun] Infinite product puzzle To: <math-fun@mailman.xmission.com> A very nice identity for analytic functions is given by (*) π cot(π z) = lim Sum_{-N <= n <= N} 1/(z-n) N—>oo or briefly: oo = "Sum 1/(z-n)" -oo The series (*) is convergent for each z in C - Z (the non-integral complex numbers) (e.g., see https://people.reed.edu/~jerry/311/cotan.pdf). Puzzle: ------- Determine which function is given by the corresponding infinite product where it converges: f(z) = lim Product_{-N <= n <= N} (1 + 1/(z-n)) N—>oo —Dan