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Power series, driving me mad

  1. May 4, 2005 #1
    summation from n=1 to infinity (x-2)^n/(n*3^n).

    My teacher got -1 <= x<= 5 as the interval of convergence because he found that x-2/3<1

    Using the ratio test i get (x-2)n/(n+1)3, consistantly. This is driving me wild. :rofl:
     
  2. jcsd
  3. May 4, 2005 #2

    Hurkyl

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    Did you remember to take the limit?
     
  4. May 4, 2005 #3
    [tex]\sum_{n=1}^{\infty} \frac{(x-2)^n}{n3^n}[/tex]

    So using the ratio test we get,

    [tex]\lim_{n\rightarrow\infty}\frac{(x-2)^{n+1}}{(n+1)3^{n+1}}*\frac{n3^n}{(x-2)^n}[/tex]

    This reduces to

    [tex]\lim_{n\rightarrow\infty}\frac{(x-2)n3^n}{(n+1)3^{n+1}}[/tex]

    Which further reduces to...

    [tex]\lim_{n\rightarrow\infty}\frac{n(x-2)}{3(n+1)}[/tex]

    So, evaluating the limit we get:

    [tex]\frac{{\mid}x-2{\mid}}{3} < 1[/tex]

    I think Hurkyl's advice was easier than my work.
     
    Last edited: May 4, 2005
  5. May 4, 2005 #4
    ah yes, thats it! i forgot the limit. thank you
     
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