Does the sequence $\{a_n\}_{n=1}^{\infty}, a_n=\sin(\lfloor 10^n \pi \rfloor)$ converge?

Question

Define a sequence $\{a_n\}_{n=1}^{\infty}$ such that $ a_n=\sin(\lfloor 10^n \pi \rfloor)$. Does this above sequence converge?

My approach was to use the $\epsilon-\delta$ definition of limits of sequences. Suppose $a_n$ converges and let its limit be $L\in\mathbb{R}$. Then $\forall \epsilon>0, \exists N \in \mathbb{N}$, such that $\forall n> N$: $|a_n-L|<\epsilon \Rightarrow L-\epsilon< a_n<\epsilon+L.$

Now, WLOG, $\epsilon<1$, and since $-1\leq a_n\leq 1$, $\epsilon +L\leq 1$ and $-1\leq L-\epsilon$ $$\Rightarrow L\leq 0.$$

This is as far as I could go.

Answer 1

Perhaps not the intended solution, assuming this is either a textbook or a competition problem, but here's a sketch...

Note that $$a_n = \sin(\lfloor 10^n \pi\rfloor) = \sin(10^n\pi - \{10^n\pi\}) = -\sin(\{10^n\pi\}).$$

Since $f : [0,1] \to \mathbb{R}$ defined by $f(x) = -\sin(x)$ is continuous and strictly decreasing, $a_n = f(\{10^n \pi\})$ is convergent if and only if $\{10^n \pi\}$ is.

By considering decimal expansions (choosing the terminating expansion when we must) and assuming $r > 0$, we can more generally show that $\{10^n r\}$ is convergent if and only if $9\cdot 10^{n_0} r$ is an integer for some natural number $n_0$. In particular, convergence of $\{10^n r\}$ implies $r$ is rational.

As $\pi$ is not rational, $\{10^n \pi\}$ is not convergent, so $a_n = f(\{10^n\pi\})$ isn't convergent.

Answer 2

In my opinion it should not. The reason for this being the following: The sine in your function approximates $\pi$ to $n$ decimals. Of note here is that $$ \tilde{a_n} := \sin(10^n *\pi ) =0 \hspace{5pt} \forall n \in \mathbb{N}. $$ Now, as $\sin$ is $2\pi$-periodic, we know that $$ \sin(\lfloor 10^n\pi \rfloor) = \sin(10^n\pi-\lfloor 10^n\pi\rfloor). $$ We can denote this by $$ b_n := 10^n \pi - \lfloor 10^n \pi \rfloor, \hspace{5pt} b_n\in (0,1) $$ which will be a zero followed by the decimals of $\pi$ starting at the $n+1$th. Note also that this sequence $b_n$ is in the open interval i.e. the boundaries are never taken.
I claim that for the original sequence to be converging, we would need this second sequence to converge which it notably does not.
If $b_n$ were converging, the decimals of $\pi$ would have to become monotonically decreasing.