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Power series.

  1. Nov 8, 2007 #1
    1. The problem statement, all variables and given/known data
    Find the radius and interval of convergence for the following power series.
    [tex]\sum_{n = 2}^{\infty}\frac {(1 + 2cos\frac {\pi n}{4})^n}{lnn}x^n[/tex]





    3. The attempt at a solution

    [tex]R = \frac {1}{\lim_{n\rightarrow\infty}\sqrt [n]{\frac {(1 + 2cos\frac {\pi n}{4})^n}{lnn}}} = \lim_{n\rightarrow\infty}\frac {e^{\frac {ln(lnn)}{n}}}{(1 + 2cos\frac {\pi n}{4})}[/tex]
    In answers [tex]R=\frac{1}{3}[/tex].
    [tex]\lim_{n\rightarrow\infty}e^{\frac {ln(lnn)}{n}} = 1[/tex].Then is

    [tex]\lim_{n\rightarrow\infty}(1 + 2cos\frac {\pi n}{4}) = 3[/tex]?

    As I know usually [tex]\lim_{x\rightarrow 0}cosx=1[/tex],not

    [tex]\lim_{x\rightarrow\infty}cosx=1[/tex]
     
  2. jcsd
  3. Nov 8, 2007 #2
    try using [tex] \frac{1}{R} =\limsup_{n \to \infty} \sqrt[n]{| a_n |}[/tex]
     
    Last edited: Nov 8, 2007
  4. Nov 8, 2007 #3

    HallsofIvy

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    cos(x) itself does not approach ANYTHING as x goes to infinity, it cycles back and forth between -1 and 1. However, at any point at which cos([itex]\pi n/4[/itex]) is 1, 2 cos([itex]\pi n/4[/itex]) is 2 and 1+ 2cos([itex]\pi n/4[/itex]) is 3. That is the largest the denominator can get.
     
  5. Nov 8, 2007 #4
    I see.Thanks!
     
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