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Sum of a series

  1. May 8, 2012 #1
    Determine whether the series is convergent or divergent. If convergent, find its sum.

    Summation (n=1 to infinity) (1+2^n) / (3^n)

    I said that it is convergent from looking at the terms. The first couple of terms, you get:

    1 + 5/9 + 9/27 + 17/108 + 32/243

    It seems as if the numbers are converging to a number.

    I don't know how to find the sum. All I know is, the sum of a series is equal to the limit of the sequence of partial sums. And from here I don't know what to do
     
  2. jcsd
  3. May 8, 2012 #2

    LCKurtz

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    You generally don't determine convergence or divergence by looking at a few terms. Think about geometric series.
     
  4. May 8, 2012 #3
    How do i determine it then? And i don't know how to move the terms around to get it to the form of a*r^(n-1)
     
  5. May 8, 2012 #4

    Mark44

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    $$\sum_{n = 1}^{\infty}\frac{1 + 2^n}{3^n} = \sum_{n = 1}^{\infty}\left(\frac{1}{3^n} + \frac{2^n}{3^n}\right)$$

    Does that help?
     
  6. May 8, 2012 #5
    I did that .. About as far as I get .. Am I supposed to put this in a geometric series form? I'm confused
     
  7. May 8, 2012 #6

    Mark44

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    Under certain conditions you can split a series that is a sum into the sum of two series. The condition is that all series involved are absolutely convergent.
     
  8. May 8, 2012 #7

    Ray Vickson

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    If you sum from n = 1 to some finite N, then you have two finite geometric series:
    [tex]\sum_{n=1}^N \left( \frac{1}{3^n} + \frac{2^n}{3^n} \right) = S_1(N) + S_2(N),[/tex]
    where
    [tex] S_1(N) = \sum_{n=1}^N \left( \frac{1}{3}\right)^n \text{ and } S_2(N) = \sum_{n=1}^N \left( \frac{2}{3}\right)^n. [/tex] There is no difficulty here in splitting up the sum into two parts, because the sums all have a finite number of terms. Each of S1(N) and S2(N) have well-defined limits as N → ∞ (given, in fact, by the geometric sum formula), so we get a nice limit for the whole sum.

    RGV
     
  9. May 9, 2012 #8
    Ok I did that too so for the (1/3)^n , a=1/3 and r=1/3? So that sum is a/(1-r) = (1/3)/(1-(1/3)) = 1/2

    And for (2/3)^n, a=2/3 and r=2/3 so the sum is (2/3)/(1-(2/3)) = 2

    And together that is 5/2 ok I got it now thanks! I always make the problems more difficult than they are
     
  10. May 9, 2012 #9

    Mark44

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    Don't do that! :smile:
     
  11. May 9, 2012 #10
    I know I wish I could stop that! By the way, so how do i find out of the series is convergent or divergent, since someone said that you can't do it by plugging in a few numbers
     
  12. May 9, 2012 #11

    micromass

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    Surely

    [tex]\sum a_n+b_n=\sum a_n + \sum b_n[/tex]

    holds for every convergent series?? As long as you don't rearrange terms, I don't see why you need absolute convergence.
     
  13. May 9, 2012 #12
    Micromass, can you explain to me how to determine whether a series is convergent or divergent? Because I just plug numbers in and see if they seem to come to a certain number
     
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