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Convergent/divergent series problem

  1. Feb 21, 2012 #1
    1. The problem statement, all variables and given/known data
    Determine whether the series is convergent of divergent. If it is convergent find its sum.

    Ʃ (1+2^n)/(3^n) as n=1 to infinite.


    2. Relevant equations



    3. The attempt at a solution
    I'm not exactly sure where to start with this. I tried to test for divergence but I don't know how to solve the limit. I tried to find a pattern with partial sums, but did not find anything relevant. Am I approaching this the wrong way? I kinda need a kick in the right direction.

    EDIT: Was looking at the limit for test of divergence incorrectly. So now I know that the limit of an = 0 (after L'Hospital's rule and whatnot). Since an is convergent, I now must find the sum of the series...
     
    Last edited: Feb 21, 2012
  2. jcsd
  3. Feb 21, 2012 #2

    vela

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    How do you know the series converges?
     
  4. Feb 21, 2012 #3
    I know the series is convergent because the limit of the series = 0. Test for divergence says that the series is divergent if the limit of the series does not exist, or if the series does not equal 0.

    I made some more progress:

    Ʃ (1 + 2^n)/(3^n)
    = Ʃ 1/(3^n) + (2^n)/(3^n)
    = Ʃ 1/(3^n) + Ʃ (2^n)/(3^n)
    = 1/2 + Ʃ (2^n)/(3^n)
    =?

    I'm not quite sure how to solve the remaining summation. Is it a geometric series?
     
  5. Feb 21, 2012 #4
    Ʃ (2^n)/(3^n) = Ʃ (2/3)^n = 2
     
  6. Feb 21, 2012 #5

    HallsofIvy

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    You may be misunderstanding. It is true that if the limit of the sequence [itex]\lim_{n\to 0} a_n[/itex] is not 0 then the series [itex]\sum_{n=0}^\infty a_n[/itex] cannot converge. But if the limit of the sequence is 0, then the series may or may not \onverge. [itex]\lim_{n\to 0} 1/n = 0[/itex] but [itex]\sum 1/n[/itex] does NOT converge.

    [tex]\frac{2^n}{3^n}= \left(\frac{2}{3}\right)[/tex].
     
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