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Telescoping Series

  1. Mar 18, 2005 #1
    I have no idea how to do this :'( really the only part I don't understand is the ending part.....like for

    the infinite sum of (1/n(n+1)).............I know you start of by partial fractions.....then you just plug in a few numbers for n.

    so I end up with:

    (1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + ... + (1/n - 1/(n+1))

    then in another problem...the infinite sum of ( 2/{(n-1)(n+1)} ) ends up like...

    (1 - 1/3) + (1/2 - 1/4) +1/3 - 1/5) + ... + (1/(n-3) - 1/(n-1)) + (1/(n-2) - 1/n)

    I want to know how you get the end results.....the ones with "n" in them....I don't understand how to get those numbers....or why they are what they are....

    I understand the first part without the "n"....but I don't know how to end it with the "n"....I hope this makes since. :confused:
  2. jcsd
  3. Mar 18, 2005 #2


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    Homework Helper

    OK, let's take a look at:
    So if we write it out longhand it turns into:
    but we can regroup:
    This is just moving the parens around, but the paired numbers nicely all zero out, leaving you with

    It should be easy to see that the pattern can be extended to any number of terms so that you end up with

    There are more formal approaches to this, but I doubt that they will be particularly helpful to you.
    Last edited: Mar 18, 2005
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