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Evaluating a limit by integral test

  1. Apr 20, 2014 #1
    1. The problem statement, all variables and given/known data

    Evaluate ## ∑^∞_{n=1} \frac {2}{n(n+2)} ##

    2. The attempt at a solution

    I've solved this question simply enough by evaluating it as a telescoping series and found the answer as 3/2. Now, when applying the integral test, it only works when dealing with positive, decreasing functions, correct? I'm not exactly sure as to why you can only apply it under circumstances though (if someone could also explain this, that would help). My question is, why can't the integral test be applied in this situation? If I'm not mistaken, you get two different answers using the two methods (i.e. integral evaluation or telescoping) yet the sum posted fulfil the aforementioned criteria.
     
  2. jcsd
  3. Apr 20, 2014 #2

    Mark44

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    Working with the summation as a telescoping series, you found the sum of the series. The integral test doesn't give you the sum of the series, it gives you the integral of the function you're integrating, and this will be reasonably close to, but not the same as the summation.

    I think that's what you're asking about, at least in part.
     
  4. Apr 20, 2014 #3

    mfb

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    Where is the problem? Is ##\frac{2}{n(n+2)}## negative or not decreasing anywhere?
    The integral gives a different value, but it confirmes that the series converges (that's all the test does).
     
  5. Apr 20, 2014 #4
    Thank you! Is there a method to determine how accurate it is?
     
  6. Apr 20, 2014 #5

    Mark44

    Staff: Mentor

    In general, no. The purpose of the integral test is to let you determine whether the series converges. In the summation, you're essentially adding the areas of a bunch of rectangles, each of width 1. The integral gives you the area under the curve y = f(x). The underlying geometric shapes for the two methods are different, which is why the two methods produce different values.
     
    Last edited: Apr 20, 2014
  7. Apr 20, 2014 #6
    In general? Do you mind expanding please? :)
     
  8. Apr 20, 2014 #7

    Mark44

    Staff: Mentor

    I didn't want to say a flat no, just in case there was some situation that I hadn't thought about. The important thing is that the integral test is just a test to determine whether a given series converges or not.
     
  9. Apr 21, 2014 #8

    micromass

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  10. Apr 21, 2014 #9

    mfb

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    The maximal error should be given by the first term, if you let the integral start at the point of the first term (here: at n=1).

    See attachment. The red dots mark the summands of the series. The red "curve" (step function) corresponds to an error-free integral, the grey "curve" to the worst case.

    attachment.php?attachmentid=68887&stc=1&d=1398070834.png
     

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