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Series with Trigonometric funtions

  1. Nov 18, 2007 #1
    [SOLVED] Series with Trigonometric funtions

    1. The problem statement, all variables and given/known data
    Determine whether the series converges conditionally,converges absolutely or diverges.

    [tex]\sum_{n=2}^{\infty}\frac{\sin (n+\frac{1}{n})}{\ln (\ln n)}[/tex]

    3. The attempt at a solution

    [tex]\frac{\sin (n+\frac{1}{n})}{\ln (\ln n)}=\frac{\sin n\cos \frac{1}{n}+\sin \frac{1}{n}\cos n}{\ln (\ln n)}=\frac{\sin n(1+O(\frac{1}{n^2}))+\cos n(\frac{1}{n}+O(\frac{1}{n^3}))}{\ln (\ln n)}[/tex]

    [tex]\sum_{n=2}^{\infty}\sin n,\sum_{n=2}^{\infty}\cos n[/tex] are bounded

    From Dirichlet's Test we can deduce that this series converges

    [tex]|a_n|=\left|\frac{\sin n(1+O(\frac{1}{n^2})}{\ln (\ln n)}\right|+\left|\frac{cos n(\frac{1}{n}+O(\frac{1}{n^3}))}{\ln (\ln n)}\right|[/tex]

    if we look for example at [tex]\left|\sum_{n=2}^{\infty}\frac{cos n}{n\ln (\ln n)}\right|>\left|\sum_{n=2}^{\infty}\frac{cos n}{n\ln (n)}\right|[/tex]

    0<|cosn|<1 and [tex]\sum_{n=2}^{\infty}\frac{1}{n\ln (n)}[/tex] diverges by Integral Test
    So this series converges conditionally!!
  2. jcsd
  3. Nov 19, 2007 #2

    Gib Z

    User Avatar
    Homework Helper

    Completely correct, nothing I can really add of value. You seem to be getting most of these right =]
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