Infinite Sums of Arctan

Date: 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/