The sequence with $n$th term is given below Consider the sequence $A_n = \sin n$ then
(a) The series is convergent as $n\to\infty$
(b) The series is divergent as $n\to\infty$
(c) $\lim_{n\to\infty} \text{ infimum} \sin(n) = 0$
(d) $\lim_{n\to\infty} \sin n = –1$
It is answer b). We have $$\limsup_{n \to \infty} \sin(n) = 1$$ and $$\liminf_{n \to \infty} \sin(n) = -1$$ Hence $$\limsup_{n \to \infty} \sin(n) \neq \limsup_{n \to \infty} \sin(n)$$ and thus the sequence $\left(\sin(n)\right)_{n \in \mathbb{N}}$ is divergent.
As a legitimation, I found the part below in Krantz, Real Analysis and Foundations:
