# Does the sequence $x_n = \sin\left([\text{first } n \text{ digits of } \pi] \cdot 10^{n - 1} \right)$ converge? [duplicate]

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?

• 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 – David Jul 31 at 16:29
• You want $\sin(3.14)$, not $\sin(314)$ (for example). – Ethan Bolker Jul 31 at 16:30
• @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? – Viktor Glombik Jul 31 at 16:38
• $x_n=\sin{n}$ is dense, as it was shown here and here. – rtybase Jul 31 at 20:03

$$\lim\limits_{n\rightarrow \infty} |a_n10^{n-1}-10^{n-1}\pi| \neq 0$$ 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
• If I understand your definition of $a_n$ correctly than $b_4 = 3,141 - \pi \approx -0,0006$ – AO1992 Jul 31 at 16:40
• In case $\pi$ contains digit zero? – Sungjin Kim Jul 31 at 17:03
• It can, but for a sufficiently large $n$ there will be a nonzero term because of $\pi$ being irrational – AO1992 Jul 31 at 17:07