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

Question

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 '19 at 16:29
@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? — ViktorStein, Jul 31 '19 at 16:38

score 3 · Accepted Answer · answered Jul 31 '19 at 16:34

3

$\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

answered Jul 31 '19 at 16:34

AO1992

1,431

If I understand your definition of $a_n$ correctly than $b_4 = 3,141 - \pi \approx -0,0006$ – AO1992 Jul 31 '19 at 16:40
In case $\pi$ contains digit zero? – Sungjin Kim Jul 31 '19 at 17:03
1

It can, but for a sufficiently large $n$ there will be a nonzero term because of $\pi$ being irrational – AO1992 Jul 31 '19 at 17:07

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

1 Answers1