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Homework Help: Series againand again

  1. Nov 9, 2004 #1
    Sorry about the title everyone but ive posted numerous threads on series and I had to choose an apropriate title :tongue2:
    The problem asks to use the ratio test, and determine for which values of x the test is conclusive-either converging or diverging. Then check those cases where the test is inconclusive by some other means.

    here is the the series [tex]\sum_{n=3}^{\infty}\frac{x^n}{n3^n}[/tex]...converge or diverge here is what i did [tex]\frac{a_{n+1}}{a_n}[/tex] and that came out to be [tex]\frac{x^{n+1}}{(n+1)(3^{n+1})}[/tex] multiplie by the [tex]\frac{n3^{n}}{x^{n}}[/tex] and after you cross out similar variables and it comes out to be

    [tex]\lim_{x\rightarrow \infty}\frac{xn}{3(n+1)}[/tex]
     
    Last edited: Nov 9, 2004
  2. jcsd
  3. Nov 9, 2004 #2

    Galileo

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    [tex]\frac{a_{n+1}}{a_n}=\frac{x^{n+1}n3^n}{x^n(n+1)3^{n+1}}=\frac{nx}{(n+1)3}[/tex]
     
  4. Nov 9, 2004 #3
    thanks galileo but I got that far just had problems latexing it
     
  5. Nov 9, 2004 #4
    So you get

    [tex]\lim_{n \to \infty} \left| \frac{n}{(n+1)}\frac{x}{3}\right| = \frac{|x|}{3}[/tex]

    And you know this is stricly less than 1 for it to be conclusively converging, strictly greater than 1 to be conclusively diverging, and inconclusive at 1. So you must test the values for which the expression equals one.

    --J
     
  6. Nov 10, 2004 #5

    James R

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    The remaining cases are x=3 and x=-3

    If x=3 then the series reduces to the harmonic series, which diverges.

    If x=-3 then we have an alternating series, and we can use the Leibnitz test (whose exact conditions escape me right now).
     
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