That looks impressive --- if the product really is constant, anyway! Not directly relevant, I know. Maple has a habit of simplifying away removable singularities that it finds, which can on occasion prove highly misleading. WFL On 10/23/18, Bill Gosper <billgosper@gmail.com> wrote:

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 _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun