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Proof for a sum to be less than an integral of the sums expression.

  1. Feb 1, 2009 #1
    1. The problem statement, all variables and given/known data

    Prove that

    [itex] \displaystyle\sum_{x=2}^n \frac{1}{x^2}<\int_0^n \frac{1}{x^2} dx [/itex]

    2. Relevant equations

    I'm not sure what equations are relevant for this, but I think for this I would need:

    [itex] \displaystyle\sum_{x}^{\infty} \frac{1}{x^s} = \displaystyle\prod_{p}^{\infty} \frac{1}{1-p^{-s}}[/itex]

    Or maybe

    [itex] \displaystyle\sum_{x}^{\infty} \frac{1}{x^2} = \frac{\pi^2}{6}[/itex]

    3. The attempt at a solution

    (1)

    [itex]\displaystyle\sum_{x=2}^{\infty} \frac{1}{x^2} = \displaystyle\sum_{x}^{\infty} \frac{1}{x^2} - 1[/itex]

    [itex]\displaystyle\sum_{x=2}^{\infty} \frac{1}{x^2} = \frac{\pi^2}{6} - 1[/itex]

    (2)

    [itex]\int_0^\infty \frac{1}{x^2} dx = \left[ -\frac{1}{x} \right]_0^\infty[/itex]

    [itex]\left[ -\frac{1}{x} \right]_0^\infty = 0[/itex] ?

    This integral does not converge between those limits?

    (3)

    [itex]\frac{\pi^2}{6} - 1 > 0[/itex]

    So it seems that I can't prove for this? Am I right in using infinity for the limits to prove for n or do I need another method? Thanks
     
    Last edited: Feb 1, 2009
  2. jcsd
  3. Feb 1, 2009 #2

    Dick

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    There's a much easier approach. Show a lower Riemann sum for the integral that equals the series. Are you sure you want the lower limit on the integral to be 0? That's divergent, you know. I'm guessing it should be 1.
     
  4. Feb 1, 2009 #3
    Yes it was an error the lower limit is not 0 it is 1.

    (1)

    [itex]\displaystyle\sum_{x=2}^{\infty} \frac{1}{x^2} = \displaystyle\sum_{x}^{\infty} \frac{1}{x^2} - 1[/itex]

    [itex]\displaystyle\sum_{x=2}^{\infty} \frac{1}{x^2} = \frac{\pi^2}{6} - 1[/itex]

    (2)

    [itex]\int_1^\infty \frac{1}{x^2} dx = \left[ -\frac{1}{x} \right]_1^\infty[/itex]

    [itex]\left[ -\frac{1}{x} \right]_1^\infty = -\frac{1}{\infty} + 1 = 1[/itex] ?


    (3)

    [itex]\frac{\pi^2}{6} - 1 < 1[/itex]

    It is proved for infinity now I think but is this enough to say that it is so for

    [itex] \displaystyle\sum_{x=2}^n \frac{1}{x^2} < \int_1^n \frac{1}{x^2} dx [/itex] ?
     
  5. Feb 1, 2009 #4

    Dick

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    Yes, the integral is 1 and pi^2/6-1 is less than one. That's fine. But what I'm suggesting is a more low tech solution in which you don't need to know that the sum of the inverse squares is pi/6. Draw the graph of 1/x^2 and fit rectangles under the curve between every pair of integers. The sum of all the rectangles is clearly less than the integral, right?
     
  6. Feb 1, 2009 #5
    Oh... I've just realised that the rectangles you're talking about actually fit that sum in the question!
     
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