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Let $\{a_n\}$ be a real sequence satisfying (1)$\displaystyle \lim_{n \to \infty} a_n=+\infty$; (2)$\displaystyle \lim_{n \to \infty}|a_n-a_{n+1}|=0.$ Prove $\{\sin a_n\}$ diverges.

Probably, we may suspect that $\{\sin a_n\}$ has more than one accumulation point. Indeed, the set of such accumulation points cover the whole interval $[-1,1]$.

There exists a theorem which could be stated as follows:

Let $\{a_n\}$ be a bounded sequence satisfying $\lim_{n \to \infty} > |a_n-a_{n+1}|=0$. Then the set of the accumulation points of $\{a_n\}$ is $\displaystyle\left[\varliminf_{n \to \infty} a_n,\varlimsup_{n \to > \infty} a_n\right]$.

This can help?

WuKong
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1 Answers1

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Let $x\in [-1,1]$ and $\epsilon > 0$ be given. By uniform continuity of $\sin$, there is a $\delta>0$ such that $$ |x - y| \lt \delta \implies|\sin x -\sin y |<\epsilon $$ Chose $N \in \mathbb{N}$ such that $|a_n - a_{n+1}| < \delta$ for all $n \ge N$. Since $a_n \longrightarrow \infty$, there exists $m,k \in \mathbb{N}$ such that $m>N$ and $$a_{m+k}> a_m + 2\pi$$ By this choice, there exists $y\in (a_m,a_{m+k})$ such that $\sin y = x$. Let $j$ be the first index in $\{0,...,k\}$ such that at $a_{m+j+1} > y$. Then we have $a_{m+j} \le y$, so $$|a_{m+j} - y| \le |a_{m+j+1} - a_{m+j}| < \delta$$ Therefore $$|\sin a_{m+j} - x| = |\sin a_{m+j} - \sin y|< \epsilon$$ Since both $x$ and $\epsilon$ were arbitrary, this proves that the values of $\sin a_n$ are dense in $[-1,1]$.

Sam
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