Bounded sequence with ratio of consecutive terms going to 1 is convergent?

Question

Let $(x_n)$ be a sequence satisfying

1. $1 \leq x_n \leq 2$ for all $n \geq 1$,
2. $\lim_{n\to\infty} x_{n+1}/x_n = 1$.

Is this enough to show that $(x_n)$ converges?

This is motivated by the case where $x_n$ is the ratio of successive Fibonacci numbers , as well as this similar example. In both cases, we can show convergence but making use of the actual definition of the sequence, not simply the two properties listed above.

I have attempted to show that this sequence is Cauchy but the details elude me.

Answer 1

Let $(x_n)_{n\geq 1}$ by

$$ x_n = \frac{3+\sin(\log n)}{2}. $$

It is obvious that $1 \leq x_n \leq 2$. Moreover,

$$ \left|\frac{x_{n+1}}{x_{n}} - 1\right| \leq \frac{1}{2}\left| \sin(\log(n+1)) - \sin(\log n)\right| \leq \frac{1}{2}\log\left(\frac{n+1}{n}\right) $$

by the mean-value theorem and this bound converges to $0$ as $n\to\infty$. So $x_{n+1}/x_n \to 1$ as $n\to\infty$. Of course, it is clear that $x_n$ does not converge as $n\to\infty$.

Answer 2

No. For example, we can take $x_n=x_{n-1}\pm\frac 1n$, and choose the signs to ensure that $1\leq x_n\leq 2$. This will also ensure the ratios tend to $1$, since $\frac{x_{n+1}}{x_n}=1\pm\frac1{nx_n}$, and $nx_n\to\infty$. We can also choose signs such that the sequence gets arbitrarily close to $1$ and to $2$ each infinitely often, since $\sum\frac1n=\infty$.