If $\{x_{n}\}_{n \geq 1}$ is a sequence of real numbers such that $\lim _{n \rightarrow \infty} \frac{x_{n}}{n}=0.001$, then
(A) $\{x_{n}\}_{n \geq 1}$ is a bounded sequence
(B)$\{x_{n}\}_{n \geq 1}$ is an unbounded sequence
(C) $\{x_{n}\}_{n \geq 1}$ is a convergent sequence
(D) $\{x_{n}\}_{n \geq 1}$ is a monotonically decreasing sequence
If $\{x_{n}\}_{n \geq 1}$ was bounded, show that $\lim _{n \rightarrow \infty} \frac{x_{n}}{n}=0$, by sandwich theorem.
If $\{x_{n}\}_{n \geq 1}$ was convergent, show that $\lim _{n \rightarrow \infty} \frac{x_{n}}{n}=0$, by algebra of limits.
If $\{x_{n}\}_{n \geq 1}$ was motonotically decreasing and bounded below, then it would have been convergent by Monotone Convergence Theorem.
Let's consider if it is not below below, i.e. $\lim _{n \rightarrow \infty} {x_{n}} = -\infty $
Find the limit in each of this case.
Hence, it will be unbounded. See the full solution and proof idea below.
