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: $$0 \le \lim_\limits{n \to \infty}\frac{\cos^2(n)}{2^n} \le 0 $$

  • $\begingroup$ This looks right, though I am not sure what you are asking? $\endgroup$
    – Norse
    Dec 16, 2019 at 13:52
  • $\begingroup$ @Norse I am asking if its right...I asked if my procedure was right? $\endgroup$ Dec 16, 2019 at 13:53
  • $\begingroup$ Oh yeah, I see now... Well, your procedure is correct $\endgroup$
    – Norse
    Dec 16, 2019 at 13:54
  • 1
    $\begingroup$ Are you sure question was about a sequence, not a series? (both converge....) $\endgroup$
    – coffeemath
    Dec 16, 2019 at 13:55
  • 1
    $\begingroup$ @coffeemath its a sequence in the book. $\endgroup$ Dec 16, 2019 at 13:59


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