Original question: Provide a sequence ${a_n}$ such that $a_n > 0, \lim\limits_{n→+∞} {a_n}^{1/n}$ exists but $\lim\limits_{n→+∞} \frac{a_{n+1}}{a_n}$ does not exist.
I figured that $\{a_n\}=\sin(n)$ would be a good candidate. Through squeeze theorem, $\lim\limits_{n→+∞} {\sin(n)}^{1/n}=1$ since $\lim\limits_{n→+∞} {-1}^{1/n}=1$ and $\lim\limits_{n→+∞} {1}^{1/n}=1$.
This now leaves me to show that $\lim\limits_{n→+∞} \frac{\sin(n+1)}{\sin(n)}$ does not exist. Using a limit calculator, I found out that this limit is -$\infty$ to +$\infty$. However, I would like to prove this limit does not exist for conciseness.
My work so far: Suppose by contradiction that $\lim\limits_{n→+∞} \frac{\sin(n+1)}{\sin(n)}=L\in\mathbb R$. Then, $\lim\limits_{n→+∞}{ \cos(1) +\cot(n)\sin(1)}=L$. From here, I'm not sure if I can do this or not, $\lim\limits_{n→+∞} \cot(n)=\frac{L-\cos(1)}{\sin(1)}$. I know this would be a contradiction, but I want to delve a bit deeper for a contradiction that is easy to see. Thus, $\lim\limits_{n→+∞} \frac{\cos(n)}{\sin(n)}=\frac{L-\cos(1)}{\sin(1)}$. From here, $\sin(n)$ or $\cos(n)$ must have a limit, which is contradictory. From that we can conclude that the original claim is true: $\lim\limits_{n→+∞} \frac{\sin(n+1)}{\sin(n)}$ does not exist.
Thank you for your time as always.