How can I show that there is a monotonically increasing sequence $\{n_j\}_{j=0}^\infty$ such that $\{\sin(n_j)\}_{j=1}^\infty$ is convergent?

Question

How can I show that there is a monotonically increasing sequence $\{n_j\}_{j=0}^\infty$ such that $\{\sin(n_j)\}_{j=1}^\infty$ is convergent?

From the convergence definition I know that if $\{\sin(n_j)\}_{j=1}^\infty$ is convergent, say $\{\sin(n_j)\}_{j=1}^\infty\to l$, then for every $\epsilon>0$, there is a natural number $N$ such that $$ n_j\geq N\Rightarrow |\sin(n_j)-l|<\epsilon $$ $|\sin(n_j)-l|<\epsilon$ can be written as $l-\epsilon<\sin(n_j)<l+\epsilon$ and so $\arcsin(l-\epsilon)<n_j<\arcsin(l+\epsilon)$ and that tells me that I need $N>\arcsin(l-\epsilon)$. If I want to show that there is a monotonically increasing sequence $\{n_j\}_{j=0}^\infty$, I need to show that $n_{j+1}>n_j$ or $n_{j+1}-n_{j}>0$.

Answer 1

The Weierstrass–Bolzano Theorem says that any bounded sequence has a convergent sub-sequence. Since $(\sin{n})$ is bounded, this implies that there exists a strictly increasing sequence of integers $(n_k)$ such that $(\sin{n_k})$ is convergent.

Alternatively, as @5xum pointed out, you can proceed constructively by letting $n_k = \pi k$ so that $\sin{n_k}= 0$ which is of course convergent.

Answer 2

Note that for a sequence to be convergent it does not require that the elements of the sequence are not all the same! For example, the sequence $(2,2, \dots)$ is convergent (try checking this with the definition).

Note that when we are required to prove an existential statement (there is some...) it suffices to give one example.

Note that the sine function is periodic with period $2\pi$. Note that $\sin (\pi/2) = 1$. So there are infinitely many $x$ such that $\sin (x) = 1$. Let $n_{j} := \frac{\pi}{2} + 2(j-1)\pi$ for all integers $j \geq 1$. Then the sequence $(n_{j})$ is strictly increasing and $(\sin(n_{j})) = (1,1,\dots)$ converges.