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Finding sum to convergent series?

  1. Feb 5, 2012 #1
    1. The problem statement, all variables and given/known data
    Decide whether convergent or divergent, if convergent, find sum.


    Ʃ as n = 1 and goes to infinity > (3^n + 2^n)/6^n

    2. Relevant equations

    a/1-r

    3. The attempt at a solution

    I'm just confused where to find the "r" to this without actually plugging in values for n, then dividing. For example, to get the answer (at least I think it's correct), I plugged 1 in for n, got 5/6, then i plugged in 2 for n, and got 11/36. I did (11/36) / (5/6), and used that for my r value, plugged that into a/1-r, and got about 1.31

    How exactly do I find the r value? Thanks.
     
  2. jcsd
  3. Feb 5, 2012 #2

    HallsofIvy

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    Try breaking it into two sums,
    [tex]\sum \frac{3^n}{6^n}+ \sum \frac{2^n}{6^n}[/tex]
    If those converge, separately, then the combination converges to the sum of those two.
     
  4. Feb 5, 2012 #3
    I respect that. Thank you. However, how do i find what the sum is?
     
  5. Feb 5, 2012 #4

    Dick

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    Write 3^n/6^n as (3/6)^n=(1/2)^n. It's a geometric series. You've probably covered those.
     
  6. Feb 5, 2012 #5
    Does this mean that r value is simply 1/2? the 2^n does not really matter, since in the long run, it becomes so miniscule?
     
  7. Feb 5, 2012 #6

    Dick

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    Of course the 2 matters, otherwise it wouldn't be (1/2)^n. What's the sum for n=1 to infinity of (1/2)^n? Review geometric series if you have to.
     
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