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Calculating the sum of a sequence

  1. May 8, 2014 #1
    1. The problem statement, all variables and given/known data
    Compute [itex]\sum\frac{4}{(-3)^n}-\frac{3}{3^n}[/itex]
    as n begins from 0 and approaches infinity


    2. Relevant equations


    3. The attempt at a solution

    I'm just getting started on sequences and series, and so far learned about the limit test, comparison test, arithmetic / geometric series and the like. Oh and a little bit about telescoping series. I'm just not sure how to approach this question. It doesn't look like a geometric or arithmetic series, and I tried to compute the first few terms to see whether it qualifies as a telescoping series, but it doesn't look like it. Thanks for any help.
     
  2. jcsd
  3. May 8, 2014 #2
    Note that ##\sum\frac{4}{(-3)^n}-\frac{3}{3^n}=\sum\frac{4}{(-3)^n}-\sum\frac{3}{3^n}=4\sum\frac{1}{(-3)^n}-3\sum\frac{1}{3^n}.## Apply geometric sum formula.
     
  4. May 9, 2014 #3
    Ah ok...so then since n begins at zero...the answer would be...
    ##4\sum\frac{1}{(-3)^n}-3\sum\frac{1}{3^n}=4*\frac{1}{1-\frac{-1}{3}}-3*\frac{1}{1-\frac{1}{3}}=4*\frac{3}{4}-3*\frac{3}{2}=\frac{-3}{2}##

    Is that right? Thanks!
     
  5. May 9, 2014 #4
    Yes. That is correct.
     
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