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I need help on a question from my homework, which asks me to find the limit of the sequence as n approaches infinity of

$$a_n = \frac{\cos^2 n}{2^n}$$

Thanks

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Numerator bounded. What does the denominator tend to? –  Host-website-on-iPage Oct 28 '12 at 8:41

3 Answers 3

up vote 2 down vote accepted

Hint: Notice $$\frac{-1}{2^n} \leq\frac{\cos^2 n}{ 2^n} \leq \frac{1}{2^n} $$ for all $n$. Now use Squeeze Rule.

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divide the problem: it's $a_n = b_n / c_n $ where $b_n = cos^2(n)$ and $c_n=2^n$. what are the limits as $n\rightarrow \infty$ of $b_n$ and $c_n$?

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The limit of $b_n$ does not exist. –  fbg Oct 28 '12 at 9:26

Using the sandwich theorem for the sequence to obtain the result of limit is $0$.

See here

http://www.youtube.com/watch?v=TpbxFJphGyg

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