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### 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)
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/
Associated Topics:
High School Calculus

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