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Homework Help: Using ratio test to find radius of convergence

  1. Dec 7, 2011 #1
    1. The problem statement, all variables and given/known data
    Ʃ((x-3)^(n)) / (n*2^(n))

    2. Relevant equations

    lim as n→ ∞ (An+1 / An)

    3. The attempt at a solution

    When dividing two fractions, invert the second and multiple to get what you see below.

    (x-3)^(n+1)/((n+1)*2^(n+1)) * (n*2^(n))/((x-3)^(n))

    Do some cross canceling to get (n(x-3)) / (2(n-1))

    Then when you take the limit as n goes to infinity, you're just left with (x-3)/2

    What I'm stuck on is what to do next. The answer is two, but I'm not seeing it.
  2. jcsd
  3. Dec 7, 2011 #2
    You use the fact that the absolute value of the ratio must be less than or equal to 1 to find x

    [itex]\left|\frac{1}{2} (-3+x)\right|<1[/itex]

    [itex]| (-3+x)|< 2[/itex]

    the radius of convergence is 2, and it is centered at x=3
  4. Dec 7, 2011 #3
    First of all, you forgot absolute value signs.

    Secondly, what does the ratio test state??
  5. Dec 7, 2011 #4
    Oh okay. I thought you had to get x by itself, so the radius of convergence would be 5.
  6. Dec 7, 2011 #5
    No, it is not 5. Read what he wrote again.
  7. Dec 7, 2011 #6
    When you get x by itself you would be solving the inequality and thus finding the interval of convergence which is different from the radius of convergence.
  8. Dec 7, 2011 #7
    I know it's not 5. That's what I assumed it was before I posted this thread. But I don't understand what the role of the -3 is.
  9. Dec 7, 2011 #8
    To make the series centered at 3.
  10. Dec 7, 2011 #9


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    It is the center point of the interval of convergence. If the radius of convergence is 2, then the series converges between 3- 2= 1 and 3+ 2= 5.
  11. Dec 7, 2011 #10
    the -3 is just the same as a graph of y = (x-3)
    in that case it centers the line at x=3

    in your case it centers the circle of convergence at x=3
  12. Dec 7, 2011 #11
    Thanks for clearing that up. But the problem only asked for radius on convergence.
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