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Compare a series sum to it's equivelent integral

  1. Sep 24, 2007 #1
    1. The problem statement, all variables and given/known data

    excuse my formatting.

    compare

    sum(from n=2 -> infinity) of 5/(n^1.1)

    to

    integral (from 1 -> infinity) of 5/(x^1.1)

    2. Relevant equations



    3. The attempt at a solution

    if it was sum and integrate from 2 it would be easy... the sum is a rienmann sum using right endpoints and delta-x of 1, and it's error will always be larger than it's integral on a decreasing curve.

    however because the function is decreasing, and the integral starts at 1, not 2, i think i have to calculate the size of the error of the rienmann compared to it's integral from 2 -> infinity, and compare that to the integral from 1 ->2.

    maybe i'm just having a brain coniption, but i cant seem to sum either the rienmann or the integral from 2 -> infinity... anyone have any suggestions?

    thanks

    stu
     
  2. jcsd
  3. Sep 24, 2007 #2
    It's easier to see what to do with a simpler function. Let's use 1/x.

    Draw a graph, using bars of width 1 to represent each value, so you end up with a bar graph of 1/x. The sum of the area under the bars is the sum of 1/x. Now notice that you can have two curves, one joining up the top left corners of the all bars, and one joining up the top right corners. These curves are 1/x curves, but one of them is shifted along by 1 unit. The area under these curves can be found by integrating 1/x, between appropriate limits. Can you see how to apply this method to your problem?
     
  4. Sep 24, 2007 #3
    it's always something simple isnt it..

    area [tex]\frac{5}{x^{1.1}} * (x - (x-1)) < \int_{x-1}^x\frac{5}{x^{1.1}}dx[/tex]

    for all values of positive x, so the sum must be smaller than the integral

    thanks for the tip

    stu
     
    Last edited: Sep 24, 2007
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