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Finding sum of convergent series.

  1. Nov 16, 2012 #1
    Hi,

    I have determined, correctly I believe, that the following series converges:
    1/[(3n-2)(3n+1)]
    Now I am asked to determine its sum. I have tried separating it into two subseries, but each time got a p-series with p=1, hence to no avail.
    The answer should be 1/3, but how may it be arrived at?
     
  2. jcsd
  3. Nov 16, 2012 #2

    Mark44

    Staff: Mentor

    I would rewrite 1/[(3n-2)(3n+1)] as the sum of two fractions, using partial fraction decomposition. When you have done that, expand the new series. You'll probably find that the series is a telescoping one.
     
  4. Nov 16, 2012 #3

    Curious3141

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    Partial fractions, following which you get a very simple telescoping series. Write out the first few terms of each.
     
  5. Nov 16, 2012 #4

    Zondrina

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    EDIT : Whoops i didn't read the whole post, my bad.
     
    Last edited: Nov 16, 2012
  6. Nov 16, 2012 #5

    Mark44

    Staff: Mentor

    The OP has already determined that the series converges. Now he/she needs to determine its sum.
     
  7. Nov 16, 2012 #6
    Thanks!
    How may I know whether the series ((-1)^n)[cos (3^n)x]^3/(3^n) converges/diverges?Should I use the Leibnitz Criterion?
    It is stated that (cos a)^3 = (1/4)(3cos a + cos 3a)
     
  8. Nov 16, 2012 #7

    SammyS

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    I suggest starting a new thread for this.
     
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