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Limit of a sequence 2

  1. May 3, 2012 #1
    1. The problem statement, all variables and given/known data
    the sequence is an = (cos(n))^2 / 2^n


    2. Relevant equations
    none really


    3. The attempt at a solution
    like i mentioned in my last post, i usually use l'hopitals or dividing by the largest exponent from the denominator. here, i dont see why i would want to use l'hopitals, so thats out of the question. i cant really divide by the largest exponent. is there a way i could break this down or something?
     
  2. jcsd
  3. May 3, 2012 #2

    Curious3141

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    What's the maximum value the numerator can take?
     
  4. May 3, 2012 #3
    1. so that means the bottom is always going to be larger.
    can i just write on my hw like ..

    (cos(n))^2 < 1, for all n
    as n approaches infinity, 2^n gets large.
    small # / large # = 0
     
  5. May 4, 2012 #4

    Curious3141

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    Yes, although I would write [itex]\cos^2 n \leq 1 \forall n \in \mathbb{N}[/itex]. The strict inequality is actually true because the only values of n that give you a value of exactly one are zero and multiples of [itex]\pi[/itex] and you're dealing with natural numbers here, but proving it is not worth the trouble, and is unnecessary here.

    So just write that, and [itex]\lim_{n \rightarrow \infty} 2^n = \infty[/itex], so the quotient tends to zero.
     
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