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
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
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.
Using the sandwich theorem for the sequence to obtain the result of limit is $0$.
See here
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 $c_n$?