Is it possible for the sequence $\{\frac{x_{n+1}}{x_n}\}$ to be unbounded but have $\lim_{n\to\infty} x_n = x$, $x_n \ne 0$

Question

Given: $$ \begin{cases} \lim_{n\to\infty} x_n = x \\ x_n \ne 0 \\ n \in \mathbb N \end{cases} $$ Is it possible for $\{\frac{x_{n+1}}{x_n}\}$ to be and unbounded sequence?

This problem comes in the context of two others which are:

1. Does $\lim_{n\to\infty}\frac{x_{n+1}}{x_n}$ exist?

Not necessarily if we for example consider: $$ x_n = \frac{sin{\pi n \over \sqrt2}}{n} $$

2. If the limit exists and is equal to $q$, prove $|q| \le 1$

This one is also easy to show using the definition of a monotone sequence.

The third part as of the question section asks whether $\{\frac{x_{n+1}}{x_n}\}$ may be unbounded, and the answer suggests that this is indeed possible, but I don't see how.

What would be such a sequence?

Answer 1

Take $$ x_n = \begin{cases} \dfrac{1}{n}, & \text{$n$ odd} \\[6pt] \dfrac{1}{n^2}, & \text{$n$ even.} \end{cases} $$

(If we assume $x \neq 0$, there is no such example.)