Let $\{a_n\}_{n\ge 1}$ be a sequence of non-zero real numbers.
Then does there exist a subsequence $\{a_{k_n}\}_{n\ge 1}$ of $\{a_n\}_{n\ge 1}$ such that either $\{\dfrac {a_{k_{n+1}}}{a_{k_n}}\}_{n\ge 1}$ converges to $1$ or $0$ or $\{\dfrac {a_{k_{n}}}{a_{k_{n+1}}}\}_{n\ge 1}$ converges to $0$ ?
We may assume that all $a_k$ are positive. If only finitely many $a_k$ are positive, we consider the sequence $\{-a_k\}_k$ instead.
If the sequence converges to a non-zero limit, we clearly find a subsequence such that the quotient converges to $1$. More generally, the same is true if the sequence has a non-zero accumulation point.
If the sequence converges to $0$ (or again more generally: if $0$ is an accumulation point), we easily find a subsequence such that the quotient converges to $0$. Just Pick $n_1=1$ and recursively $n_{k+1}=\min\{\,i\in\Bbb N\mid i>n_k, 0<a_i<\frac1ka_{n_k}\,\}$.
Remains the case that the sequence has no accumulation point at all. But then it is unbounded and $0$ is an accumulation point of $\frac1{a_n}$. A subsequence of this with limit quotient $0$ corresponds to a limit quotient $+\infty$ of the original sequence.