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Simple Geometric series

  1. Mar 6, 2009 #1
    1. The problem statement, all variables and given/known data
    actually got two questions but both are related so put them in the same place
    the question asks
    Determine whether the geometric series is convergent or divergent. If it is convergent, find its sum.
    Inf
    1.) E 6(0.9)^(n-1)
    n=1

    Inf (-3)^(n-1)
    2.) E ---------------
    n=1 4^(n)

    the (E) is the sigma sign asking for sum
    the Inf is infinity and n = 1 is inital starting point
    they want us to evaluate the series sum from 1 to infinity

    3. The attempt at a solution

    not really sure where to start. i can figure out the converging or diverging part just plug in some numbers and see if it is getting bigger (going to infinity) or if it leavels off.
    not sure how to find the sum.
    probaly just a simple formula but the Professor did not give it to us.
     
  2. jcsd
  3. Mar 6, 2009 #2
    sorry the format was lost when posted
    the second problem is a division problem the -3 part on top
    and the 4^n on bottom that is why the lines are there.
     
  4. Mar 6, 2009 #3
    Last edited by a moderator: Apr 24, 2017
  5. Mar 6, 2009 #4
    ok that would make it a lot easier
    but how do you find "r" the ratio?
    for example number two a(1) = 1/4
    a(2) = -3/16 and a(3) = 9/64
    dont know what you would do to 1/4 to get -3/16 and you have to do that same thing to -3/16 to get 9/64 right??
     
  6. Mar 6, 2009 #5
    You are looking for a number r such that a(1)*r=a(2), and since the series is geometric, you will also have a(2)*r=a(3) for the same r. So what is r?
     
  7. Mar 6, 2009 #6
    man im dumb lol i solve the first on the wrong way. got -4/3 and then -3/4 for the second that is where i messed up.
    so my r would be -3/4
    thanks now i can just follow the formula
     
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