WTP: $\lim_{n \to \infty}\left |\frac{a_{n+1}}{a_n} \right | = r$ implies $\limsup_{n\to \infty} |a_n|^{1/n} = r$, where $n$ belongs to the naturals and $r$ is a real number.
Here is my thinking so far:
Start with the definition of convergence - so for all $\epsilon > 0$ there exists natural $N$ st. if $n>N$ then $\left| \left|\frac{a_{n+1}}{a_n} -r\right| \right|<\epsilon$ and so with a little work we can bound $a_{n+1}: |a_n|(r-\epsilon)<|a_{n+1}|<|a_n|(r+\epsilon)$...
That's all I have so far...
Any help would be appreciated!
Hint: for any sequence $(a_n)$ of positive numbers we have
$$ \liminf_{n\to\infty} \frac{a_{n+1}}{a_n} \le \liminf_{n\to\infty} \sqrt[n]{a_n} $$ and
$$ \limsup_{n\to\infty} \sqrt[n]{a_n} \le \limsup_{n\to\infty} \frac{a_{n+1}}{a_n}. $$
If we know that $\lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n}\right|$ exists and is finite then what can we say about its limits supremum and infimum?