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Radius / Interval of Convergence (Power Series)

  1. Nov 27, 2006 #1
    I need help finding the radius & interval of convergence of the following power series:

    The sum from n=0 to infinity of...

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

    I think the ratio test is supposed to work, but I narrow the limit of the test down to 1/|x-1|^3 < 1, and this doesn't make sense to me. Is there an easier way, or am I doing something wrong?

    Thanks.
     
  2. jcsd
  3. Nov 27, 2006 #2
    [tex] \sum_{n=0}^{\infty} (2-n^{\frac{1}{2}})(x-1)^{3n} [/tex]

    Is this the question?
     
    Last edited: Nov 27, 2006
  4. Nov 27, 2006 #3
    Yeah, that's the one. Is the ratio test the right approach? Is 1/|x-1|^3 < 1 the right result? I'm just not sure where to go from there, or even if that is right for the limit. Thanks a lot for your help.
     
  5. Nov 27, 2006 #4
    Oh except that only the "n" is raised to 1/2. Not (2-n). The quantity is (2-n^.5)
     
  6. Nov 27, 2006 #5
    [tex] \sum_{n=0}^{\infty} (2-n^{\frac{1}{2}})(x-1)^{3n} [/tex]

    Now that's right. Interval/Radius of convergence is the issue. I just don't know what I'm doing wrong with the ratio test.

    Thanks.
     
  7. Nov 27, 2006 #6
    Using the ratio test:

    [tex] (x-1)^{3}\; \frac{2-(n+1)^{\frac{1}{2}}}{2-n^{\frac{1}{2}}} [/tex] As [tex] n\rightarrow \infty [/tex] we get [tex] |(x-1)^{3}| < 1 [/tex]
     
  8. Nov 27, 2006 #7
    I appreciate it. Thanks!!
     
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