Given the Sequence:

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

I went on Symbolab to check what the answer was because I originally thought it was divergent because the $$\cos^2(n)$$ is bounded as follows $$0 \le \cos^2(x) \le 1$$. Symbolab hinted to use squeeze theorem, and I was wondering if my procedure was correct?

$$\lim_\limits{n \to \infty} (\frac{\cos^2(n)}{2^n})$$ Applying Squeeze theorem I did the following: $$0 \le \cos^2(n) \le 1$$ $$\frac{0}{2^n}\le \frac{\cos^2(n)}{2^n} \le \frac{1}{2^n}$$ Then I took the limit as $$n \to \infty$$. $$\lim_\limits{n \to \infty}\frac{0}{2^n} \le \lim_\limits{n \to \infty}\frac{\cos^2(n)}{2^n} \le \lim_\limits{n \to \infty} \frac{1}{2^n}$$ $$\therefore$$ one can say: $$0 \le \lim_\limits{n \to \infty}\frac{\cos^2(n)}{2^n} \le 0$$

• This looks right, though I am not sure what you are asking? Dec 16, 2019 at 13:52
• @Norse I am asking if its right...I asked if my procedure was right? Dec 16, 2019 at 13:53
• Oh yeah, I see now... Well, your procedure is correct Dec 16, 2019 at 13:54
• Are you sure question was about a sequence, not a series? (both converge....) Dec 16, 2019 at 13:55
• @coffeemath its a sequence in the book. Dec 16, 2019 at 13:59