Infinite Sums of ArctanDate: 04/15/2002 at 16:32:04 From: Greg Garcia Subject: Infinite sums of arctan The sum as n=0 to n=infinity of arctan ((1)/(n^2 +n +1)): I know it equals pi/2 but I don't see how. I have attempted to use partial fractions, but I can't factor the denominator. I konw that the arctan of 1 is pi/4 + the sum of all the angles smaller than pi/4 will eventually = pi/2 but I was hoping for something a little more calculus-based, maybe using telescoping series or something like that. Date: 04/15/2002 at 16:47:21 From: Doctor Anthony Subject: Re: Infinite sums of arctan If tan(-1)(n+1) = A tan^(-1)(n) = B n+1 = tan(A) n = tan(B) tan(A) - tan(B) n+1 - n 1 tan(A-B) = ------------------ = ----------- = -------- 1 + tan(A).tan(B) 1 + n(n+1) n^2+n+1 A-B = tan^(-1)[1/(n^2+n+1)] tan^(-1)[n+1] - tan^(-1)[n] = tan^(-1)[1/(n^2+n+1)] So we can set up a series of equations u(0) = tan^(-1)(1) - tan^(-1)(0) u(1) = tan^(-1)(2) - tan^(-1)(1) u(2) = tan^(-1)(3) - tan^(-1)(2) u(3) = tan^(-1)(4) - tan^(-1)(3) u(4) = tan^(-1)(5) - tan^(-1)(4) ................................ ................................ u(n) = tan^(-1)(n+1) - tan^(-1)(n) ------------------------------------ add all the equations SUM[u(r)] = tan^(-1)(n+1) - tan^(-1)(0) all other equations on the right cancel between rows. inf SUM[u(r)] = tan^(-1)[inf] - tan^(-1)(0) r=1 = pi/2 - 0 = pi/2 - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
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