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Sum of infinite series

  1. Apr 7, 2010 #1
    1. The problem statement, all variables and given/known data

    Given the Infinite Series 1/(1+n^2) where n goes from 1 to infinity, show that the sum is less than pi/2.

    2. Relevant equations


    Series goes 1/2, 1/5, 1/10, 1/17, 1/26 and so on

    3. The attempt at a solution

    I have tried to find a telescoping series, but I can't see to get the terms to cancel out. My next try was to find the partial sum of the series, but I seem to want to take the integral from 1 til n+1 (as a form of partial sum) of 1/(1+n^2). I end up with Arctan(1+n) - Arctan(1), which obviously is less than pi/2, but I don't find this as a credibal solution....

    Could anyone try to give me any hints on which way to go, or what way to go?

    Thanks for welcoming me to the forum!
  2. jcsd
  3. Apr 7, 2010 #2
    This is nonsense; perhaps you mean [tex]\int \frac{dx}{1 + x^2} = \arctan x + C[/tex] ?

    You are on the right track, actually. One small problem is that you need to overestimate the "bar graph" of the sum of the series with the curve, and [tex]\frac1{1 + x^2}[/tex] is a decreasing function of [tex]x[/tex], which means you need to move the "bar graph" left to fit under the curve. That is, for the [tex]n = 1[/tex] term, you need to write [tex]\frac1{1 + 1^2} < \int_0^1 \frac{dx}{1 + x^2}[/tex] and not [tex]\frac1{1 + 1^2} < \int_1^2 \frac{dx}{1 + x^2}[/tex] (which is false). If what I said about bar graphs doesn't make sense, draw a few pictures.
  4. Apr 8, 2010 #3
    2. I tried to write the integralsign, but couldnt find it so I did it the easy way...

    I totally see it now, and my problem is solved. Thanks alot for helping me! This is a wonderful forum!

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