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Limit of a sequence

  1. Feb 1, 2008 #1
    1. The problem statement, all variables and given/known data
    Determine whether the sequence converges or diverges. If it converges, what does it converge to?

    (cos[n])^2 / (2^n)

    2. Relevant equations
    L'Hospital's Rule
    or
    Squeeze Theorem


    3. The attempt at a solution
    As n-->infinity, this function approaches infinity/infinity. Applying L'Hospital's rule gives -(sin[n])^2 / 2 which gives infinity/2=infinity.

    However, if I use the squeeze theorem to say that 0<(cos[n])^2<1 and then divide everything by 2^n, then I can say that this function approaches 0.

    Which is the correct method?
     
  2. jcsd
  3. Feb 1, 2008 #2

    Vid

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    The derivative of [tex]2^{n} = ln(2)*2^{n}[/tex], and the derivative of [tex]cos^{2}(n) = -2cos(n)sin(n).[/tex]
     
    Last edited: Feb 1, 2008
  4. Feb 1, 2008 #3

    Dick

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    Science Advisor
    Homework Helper

    i) The sequence isn't infinity/infinity. You just said 0<=cos(n)^2<=1. How can it be infinity/infinity?? So you can't use l'Hopital. 2) The derivative of 2^n is NOT 2. The squeeze argument is correct.
     
  5. Feb 1, 2008 #4

    HallsofIvy

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    Staff Emeritus
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    I would also point out that [itex]\lim_{n\rightarrow \infty} sin(n)[/itex] is NOT infinity. It just doesn't exist.
     
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