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Sums of Series

  1. Jul 27, 2008 #1
    [tex]\sum[/tex] as n=0(theres a infinity above that sigma) , 3(1/11)^n

    I thought it would just be 3/11 and converge due to the geometric test but its not....to find the sum would I just start at 0 and put numbers in for "n"...
  2. jcsd
  3. Jul 27, 2008 #2


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    You mean [tex]\sum^{\infty}_{n=0} 3\left(\frac{1}{11^n}\right)[/tex]. You can factor the 3 outside of the sigma. And the resulting would be a geometric series, no?
  4. Jul 27, 2008 #3
    yes but I dont see how I would go about finding a actual sum.
  5. Jul 27, 2008 #4


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    What do you know about the sum of a geometric series?
  6. Jul 27, 2008 #5
    OH! S=a/1-r
  7. Jul 27, 2008 #6
    its 33/10 thankyou
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