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$\{a_n\}$ is a sequence such that $a_n=\cos(n)$ with $n∈Z^+$. What I want is to prove that $\max(\cos(n))\to1$ (and also $\min(\cos(n))\to-1$). I thought I could choose $n=\lfloor{\pi\times2\times10^M}\rfloor$ with $M∈Z^+ $ and that $\lim_{M\to\infty} \cos(\lfloor{\pi\times2\times10^M}\rfloor)=\lim_{M\to\infty}\cos(\pi\times2\times10^M)=1$ would hold, but I now realize that $\lim_{M\to\infty}\{\pi\times2\times10^M\}$ does not converge to $0$, making my attempt incorrect. Am I even correct in assuming that $\max(\cos(n))\to1$?

HideBehind
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  • I used $\cos(n)$ in place of $a_n$ as the former is the definition of the latter, so $\max(\cos(n))$ is just $\max(a_n)$. ($\max()$ refers to the maximum value whatever inside of the parenthesis can get) – HideBehind Dec 23 '23 at 21:01
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    Depending on what you mean by $Z^+$ this could be trivial because $cos(0)=1$. – Michael Dec 23 '23 at 21:09
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    Your max notation is off, I think you mean $$h(n)=\max{\cos(i) : 1\leq i \leq n}$$ and you want to show $\lim_{n\rightarrow\infty} h(n)=1$. – Michael Dec 23 '23 at 21:10
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    Yes, your assumption that $\sup { \cos n \mid n \in \Bbb Z^+ }=1 $and $\inf { \cos n \mid n \in \Bbb Z^+ } = -1$ is correct. The set of all $\cos n$ is in fact dense in $[0, 1]$. https://math.stackexchange.com/questions/656930/show-that-a-set-is-dense-in-1-1?noredirect=1&lq=1 – Robert Shore Dec 23 '23 at 21:13
  • @Michael $Z^+$ refers to positive integers here so $n=0$ is not an answer sadly. I'm also unsure about how the max notation should be shown, so you might be correct. In any case, my goal is to find the lowest number that's bigger than or equal to every element of the sequence ${a_n}$ – HideBehind Dec 23 '23 at 21:13
  • You want to find integers $n$ for which $n \approx 2\pi k$ for some integer $k$. Consider rational approximations of $\pi$. – Michael Dec 23 '23 at 21:17
  • @Michael the problem with that approach is that it's not possible to get rid of the digits after the decimal point of 2. – HideBehind Dec 23 '23 at 21:24
  • @RobertShore's answer can be reached by following the replies to the question this post is a duplicate of. – HideBehind Dec 23 '23 at 21:35

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