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I was trying to come up with an interesting example (not like $(-1)^n$ or cyclic like $1,2,3, 1,2,3, 1,2,3, \ldots$) of a bounded not converging sequence with an explicit convergent subsequence.

I though of something like $x_n := \sin(n)$. Since $\sin(k\pi) = 0$ for every $k \in \mathbb{Z}$ I thought that the closer the argument of the sine was to a multiple of $\pi$, the closer sine of that argument should be to zero (continuity...). Therefore, the sequence $x_n := \sin(a_n \cdot 10^{n - 1})$, where $a_n$ are the first $n$ digits of $\pi$ should be converging to $0$, right? If yes, how can I show it?

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    $\begingroup$ I don't see why this sequence should even converge at all. You are not getting closer and closer to multiples of $\pi$, as you may be expecting $\endgroup$ – David Jul 31 at 16:29
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    $\begingroup$ You want $\sin(3.14)$, not $\sin(314)$ (for example). $\endgroup$ – Ethan Bolker Jul 31 at 16:30
  • $\begingroup$ @David But $a_n \to \pi$, so $\sin(a_n) \to \sin(\pi) = 0$, the only problem is that $a_n 10^{n - 1} \not\to \pi$, right? $\endgroup$ – Viktor Glombik Jul 31 at 16:38
  • $\begingroup$ $x_n=\sin{n}$ is dense, as it was shown here and here. $\endgroup$ – rtybase Jul 31 at 20:03
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$\lim\limits_{n\rightarrow \infty} |a_n10^{n-1}-10^{n-1}\pi| \neq 0 $ because for any large $N$ you can find $n>N$ such that this difference will be bigger $0.1$.

You are looking for $a_n-\pi$ in this case you will have monotonic convergence

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  • $\begingroup$ If I understand your definition of $a_n$ correctly than $b_4 = 3,141 - \pi \approx -0,0006$ $\endgroup$ – AO1992 Jul 31 at 16:40
  • $\begingroup$ In case $\pi$ contains digit zero? $\endgroup$ – Sungjin Kim Jul 31 at 17:03
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    $\begingroup$ It can, but for a sufficiently large $n$ there will be a nonzero term because of $\pi$ being irrational $\endgroup$ – AO1992 Jul 31 at 17:07

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