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Find the sum of the series

  1. Aug 19, 2008 #1
    1. The problem statement, all variables and given/known data:
    The following is a geometric series.
    Determine whether series is converges or not.
    For the series which converge, enter the sum of the series

    2. The attempt at a solution:
    I've looked into calculating [tex]{r}=\frac{a_{n+1}}{a_{n}}[/tex] but the series isn't constant and neither is r : [tex]\frac{10}{8}{+}\frac{58}{64}{+}\frac{185}{256}{+...}[/tex]

    I feel like there is another approach to solving this problem that I am missing. Your help is appreciated, thanks.
  2. jcsd
  3. Aug 19, 2008 #2


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    Well it's just the sum of two geometric series isn't it?
    \sum^{\infty}_{n=1}(\frac{7}{8})^n + \sum^{\infty}_{n=1}(\frac{3}{8})^n
  4. Aug 20, 2008 #3
    I ran those numbers through on paper before too, but it never clicked that I could find the sum individually and add them together. Thanks for your help.
  5. Aug 24, 2008 #4
    does it matter that the series is 1-relative? i.e. does one need to convert to make r^(n-1)?

  6. Aug 24, 2008 #5


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    Good point.
    [tex]\sum_{n=0}^\infty a r^n= \frac{a}{1- r}[/tex]
    Here, since it is missing the "n= 0" term, you just subtract that term of: ar0= a so
    [tex]\sum_{n=1}^\infty a r^n= \frac{a}{1- r}- a= \frac{a}{1-r}- \frac{a- ar}{1-r}= \frac{ar}{1- r}[/tex]

    Of course, that is exactly the same as if you had just factored out an "r":
    [tex]\sum_{n=1}^\infty ar^n= \sum_{n= 1}^\infty (ar)r^{n-1}= \sum_{j= 0}^\infty (ar)r^j[/itex]
    where j= n-1 so that when n= 1, j= 1-1= 0.
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