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I'm trying to prove if it exists some sequence $x_n\in\mathbb{N}$ such that $\cos(x_n)\to 1$.

I've considered $x_n= 2\lfloor \pi\ 10^{n-1} \rfloor$, which is $x_1=2\cdot 3, x_2=2\cdot 31, x_3=2\cdot 314, x_4=2\cdot 3141$ and so on.

I don't think this initial approach verifies $\cos(x_n)\to 1$. Maybe it doesn't exist.

Any help would be appreciated.

1 Answers1

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Yes, such a sequence exists. Define $x_n$ as follows. Consider the sequence $\{2\pi\}, \{4\pi\}, \{6\pi\},\ldots$ (where $\{x\}$ is the fractional part of $x$). As this is a list of numbers is the interval $[0,1)$, by the pigeonhole principle we can find two positive integers, say $m_1 < m_2$ such that $|\{2m_2\pi\}-\{2m_1\pi\}|<1/n$, hence $2(m_2-m_1)\pi\in (x_n-1/n,x_n+1/n)$ for some $x_n\in \mathbb{N}$.

Set $y_n=x_n-2(m_2-m_1)\pi$, then $y_n\in (-1/n,1/n) \quad \forall n\in \mathbb{N}$. By periodicity of $\cos$, $\cos(x_n)=\cos(y_n)\to 1$.

user6725906
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  • But $n=n(m_1, m_2)$, you can't define $y_n$ in function of $m_1$ and $m_2$ –  Mar 15 '21 at 08:43
  • Sorry, I'm not sure what you mean by this. The idea of my solution is for each $n$ to find a multiple of $2\pi$ that is closer to an integer than $1/n$. Setting this integer to be $x_n$, $x_n \to 0 \pmod{2\pi}$, so $\cos(x_n)\to 1$. – user6725906 Mar 15 '21 at 12:28