The warm-up product is very beautiful; I've never seen it before. I can use the fact that (1/2)*log|(1+x)/(1-x)| = x + x^3/3 + x^5/5 + x^7/7 + ... for |x| < 1 to transform log(coth(pi/2)) + log(coth(3*pi/2) + log(coth(5*pi/2) + ... to csch(pi) + csch(3*pi)/3 + csch(5*pi)/5 + ... . So it remains to prove that the last series is equal to log(2)/8. This probably could be done with residue calculus, but I haven't tried using it to sum anything like this before. So, that's it for my attempt at the "easy" warm-up. Warut On Sun, Jan 24, 2010 at 12:58 PM, Bill Gosper <billgosper@gmail.com> wrote:
had among its terminal nodes In[652]:= Sum[(-1)^n*ArcTan[Csch[Pi*(n - 1/2)/Sqrt[3]]], {n, \[Infinity]}] == -5*Pi/24 Out[652]= 1 (n - -) Pi n 2 Sum[(-1) ArcTan[Csch[-------------]], n >= 1] Sqrt[3] 5 Pi = - ----. 24 It doesn't *look* that hard ... Joerg? Easy warmup: In[654]:= Product[Coth[Pi*(n-1/2)]^8, {n, \[Infinity]}] == 2 Out[654]= 1 8 Product[Coth[(n - -) Pi] , n >= 1] = 2. 2 --rwg _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun