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Homework Help: Series Convergence/Divergence Problem

  1. Feb 15, 2010 #1
    1. The problem statement, all variables and given/known data

    Does the series converge or diverge? Give reason for your answer; if it converges, find its sum.

    [tex]\infty\sum[/tex] n=0 [tex]\frac{(-1)^{(n+1)} * 3 - 1}{2^n}[/tex]

    2. Relevant equations

    If |r|<1, the geometric series converges to a/(1-r). If |r|> or = 1, it diverges.

    3. The attempt at a solution

    [tex]\infty\sum[/tex] n=1 [tex]\frac{(-1)^{(n+1)} * 3}{2^n}[/tex] <---- This is a very similar problem that I was able to figure out.

    In this problem, it's a geometric series that converges to 1 (with a sum of [tex]\frac{(3/2)}{1-(-1/2)}[/tex].

    However, this particular problem has two differences. One is the -1 on top, which shouldn't matter as n-> infinity since it's so small. The other difference is it starts at zero, so I'm not sure if the above equation (in part "b") is relevant, with a/(1-r). I tried finding an "a" and got -4, but I'm not sure what to do about the r, besides assuming that the 3 and -1 don't matter:

    Then I'd get (-1)^n+1 / 2^n which would be 1/2 = r. Help?
     
    Last edited: Feb 15, 2010
  2. jcsd
  3. Feb 15, 2010 #2

    Mark44

    Staff: Mentor

    Fixed your LaTeX. You can see what I did by double-clicking the summations.
     
  4. Feb 15, 2010 #3
    Ah, thank you!
     
  5. Feb 15, 2010 #4

    Mark44

    Staff: Mentor

    I would expand the summation to see how close it is to the summation you've already worked with.
     
  6. Feb 15, 2010 #5
    In terms of the series itself, the ACTUAL series I'm working with (with the extra -1) gets closer to zero more slowly than the other.

    I can't use the comparison, ratio/root, or integral tests on this either. Only the geometric series formula.

    So, in other words...

    The top term oscillates between 2 and -4 and the bottom goes to infinity, thus Lim (n -> infinity) = 0 so the series converges to SOME number, correct so far?
     
    Last edited: Feb 15, 2010
  7. Feb 15, 2010 #6

    Mark44

    Staff: Mentor

    If you expand the series you'll see that it is a geometric series. What do the first four or five terms of this series look like?
     
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