Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

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


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

share|cite|improve this answer

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 $c_n$?

share|cite|improve this answer
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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.