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A Question about Power Series

  1. Oct 25, 2012 #1
    If we have the series [itex]\sum_{n=0}^{\infty} 2^{-n}x^{3n}[/itex]. We need to calculate the radius of convergence...the textbook says that, since the power of x is 3n instead of 3, we need to rewrite it in the form [itex]\sum_{n=0}^{\infty} a^{n}x^{n}[/itex]..and we say that [itex]a_{3k}=2^{-k}[/itex]. I'm not sure if I understand this...so are they setting n=3k? But we know that the power of 2 is -n, so if n=3k then doesn't the power of 2 need to be -3k?

    Thanks in advance
  2. jcsd
  3. Oct 25, 2012 #2

    Bah. Apply directly D'Alembert's (quotient or ratio) test with [itex]\,a_n=2^{-n}x^{3n}\,[/itex]:

    $$\left|\frac{a_{n+1}}{a_n}\right|=\left|\frac{x^{3n+3}}{2^{n+1}} \cdot \frac{2^n}{x^{3n}}\right|=\frac{|x|^3}{2} \xrightarrow [n\to\infty] {}\frac{|x|^3}{2}<1\Longleftrightarrow |x|<\sqrt [3] 2$$

    and there you have your convergence radius. There's only left to know whether either of the convergence interval's extreme

    points are contained in it.

  4. Oct 25, 2012 #3


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    No. They are thinking of [itex]x^{3n}[/itex] as [itex](x^3)^n[/itex] so that they will get the radius of convergence in terms of [itex]x^3[/itex] rather than just x. You can then take the cube root.

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