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Ratio Test

  1. Jan 30, 2012 #1
    [tex]\displaystyle\sum_{n = 1}^{\infty}\frac{(n!)^3}{(3n)!}z^n , \ z\in\mathbb{C}[/tex]

    By the ratio test,
    [tex]
    \displaystyle L = \lim_{n\to\infty}\left|\frac{[(n + 1)!]^3 z^{n + 1} (3n)!}{[3(n + 1)]! (n!)^3 z^n}\right| = \lim_{n\to\infty}\left|\frac{z (n + 1)^2}{3}\right| = \infty.[/tex]

    Therefore, the series diverges and there is no radius of convergence.

    Correct?
     
    Last edited: Jan 30, 2012
  2. jcsd
  3. Jan 30, 2012 #2

    lanedance

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    i don't think your cancellation is quite correct 3(n+1)=3n+3
     
  4. Jan 30, 2012 #3
    Thanks, I see the problem.

    [tex]|z|<27[/tex]

    So [itex]R = 27[/itex] then
     
    Last edited: Jan 30, 2012
  5. Jan 30, 2012 #4

    lanedance

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    can you show how you got there?
     
    Last edited: Jan 30, 2012
  6. Jan 30, 2012 #5
    [tex]\lim_{n\to\infty}\left|\frac{z (n + 1)^3}{(3n + 3)(3n + 2)(3n + 1)}\right|[/tex]


    [tex]\Rightarrow \lim_{n\to\infty}|z|\left(\frac{n^3}{27n^3}\right) = \frac{1}{27}|z| < 1 \Rightarrow |z| < 27[/tex]
     
  7. Jan 30, 2012 #6

    lanedance

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    yeah looks good

    you can just just take the result below if you like , rather than carrying the z through though its good to understand why
    [tex] r = \lim_{n \to \infty} \left|\frac{c_n}{c_{n+1}}\right| [/tex]
     
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