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Determine Convergence or Divergence. If conv. find the sum:

  1. May 5, 2012 #1
    1. The problem statement, all variables and given/known data
    ##\sum \dfrac {1+2^{n}} {3^{n}}##

    According to Wolfram Alpha the sum is 5/2. But, I think that my method is fine and shows another result.

    3. The attempt at a solution
    ##\sum \dfrac {1+2^{n}} {3^{n}}=\sum \left[ \left( \dfrac {1} {3}\right) ^{n}+\left( \dfrac {2} {3}\right) ^{n}\right]##

    Now, by geometric series property and property of infinite series:
    ##\dfrac {1} {1-\dfrac {1} {3}}+\dfrac {1} {1-\dfrac {2} {3}}=\dfrac {3} {2}+3## =9/2 =4.5

    Also, I found by the ratio test that it is absolutely convergent. (2/3) < 1

    I don't know, why Wolfram Alpha shows that the final sum is 5/2.

    Thank you.
     
    Last edited: May 5, 2012
  2. jcsd
  3. May 5, 2012 #2

    LCKurtz

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    You didn't show where the index of summation starts. Is it ##n=1## or ##n=0\ ##? It affects the first term and the sum.
     
  4. May 5, 2012 #3
    Sorry, the sum goes from n=1 to infinity.
     
  5. May 5, 2012 #4

    LCKurtz

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    So, does that answer your question?
     
  6. May 5, 2012 #5
    Ok. I still don't see it.
    S1= 3/3 =1
    S2= 5/9
    S3=9/27
    S4=17/81

    I just checked my textbook and Geometric series start from n=0. This is what you are referring to.
    Ok, so I cannot use the Geometric series method. How can I find a solution?
     
  7. May 5, 2012 #6

    LCKurtz

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    If the first term of a geometric series is ##a## and the common ration is ##r## then the sum of the series is ##\frac a {1-r}##. If you phrase it that way, it doesn't matter how it is indexed.
     
  8. May 5, 2012 #7
    Ok. So, what you are saying is that my 1st term is equal to a. Then, a=1. I can't determine the ratio from the summation. Could you help?

    I want to correct my previous statement for the sums:
    S1=1
    S2=1+5/9=14/9
    S3=(9/27)+14/9=51/27
    S4=17/81 +51/27 = 170/80
    .
    .
    . so on
     
  9. May 5, 2012 #8
    Since the index starts at 1, that means the first terms of the series are 1/3 and 2/3 respectively. And with a representing the first term of each from what LCKurtz said,[tex]\sum^\infty_{n=1} \left[ \left( \dfrac {1} {3}\right) ^{n}+\left( \dfrac {2} {3}\right) ^{n}\right] = \dfrac {\frac{1}{3}} {1-\dfrac {1} {3}}+\dfrac {\frac{2}{3}} {1-\dfrac {2} {3}}[/tex]
     
  10. May 5, 2012 #9
    Wow, so in this case. a is not only the coefficient, but also the ratio.

    Ok, thank you. I like to understand.
     
  11. May 5, 2012 #10

    Ray Vickson

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    Of course you can use the geometric series method; you just have to subtract the n=0 terms from what you wrote. That will give you the sum from n=1 to ∞. Alternatively, you can recognize that for |r| < 1 you have [itex] r + r^2 + r^3 + \cdots = r(1 +r + r^2 + \cdots),[/itex] and use your geometric sum result for the quantity in brackets.

    RGV
     
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