1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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

  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


    User Avatar
    Science Advisor

    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!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook