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I have seen a proof of $\displaystyle\limsup_{n\to\infty}|\cos{n}|=1$ by using density of $\{a+b\alpha: a,b\in \mathbb{Z}\}$ in $R$, where $\alpha$ is irrational.

Here I give another proof of as following:

See this article, a special case is that there is two increasing sequences of odd positive integers $(p_n),(q_n)$ such that $$ \left|\pi - \frac{p_n}{q_n} \right| \leq \frac{1}{q_n^2},\quad n>1.$$

Note that $|\cos (\pi-x) |= |\cos x |$ for $x\in [0,\pi]$, then $$|\cos \left(q_n\pi - p_n\right)|= |\cos p_n| \geq \cos\frac{1}{q_n} \to 1.$$ therefore $|\cos p_n|\to 1$.

Is this solution right? Thanks in advance.

quasi
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Jacob.Lee
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1 Answers1

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Your proposed alternate proof looks good, but for clarity, I would provide a little more detail:$\\[6pt]$ \begin{align*} & \left|\pi-\frac{p_n}{q_n} \right|\le \frac{1}{q_n^2}\\[4pt] \implies\;& \left|q_n\pi-p_n\right|\le\frac{1}{q_n}\\[4pt] \implies\;& 0 \le \left|q_n\pi-p_n\right|\le\frac{1}{q_n} < \pi\\[4pt] \implies\;& \cos\left(\left|q_n\pi-p_n\right|\right)\ge\cos\Bigl(\frac{1}{q_n}\Bigr)\\[4pt] \implies\;& \cos\left(q_n\pi-p_n\right)\ge\cos\Bigl(\frac{1}{q_n}\Bigr)\\[4pt] \implies\;& \cos\left(\pi-p_n\right)\ge\cos\Bigl(\frac{1}{q_n}\Bigr) &&\text{[since $q_n$ is odd]} \\[4pt] \implies\;& -\cos(p_n)\ge\cos\Bigl(\frac{1}{q_n}\Bigr)\\[4pt] \implies\;& \cos(p_n)\le -\cos\Bigl(\frac{1}{q_n}\Bigr)\\[4pt] \implies\;& \lim_{n\to\infty} \cos(p_n)=-1 &&\text{[since the sequence $(q_n)$ is increasing]} \\[4pt] \implies\;& \lim_{n\to\infty} |\cos(p_n)|=1 \\[4pt] \end{align*}

quasi
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