If $a_n$ is a sequence of real positive numbers, is it true that $\textrm{limsup}_{n \to \infty} \sqrt[n]{a_n}$ is finite exactly when you can find a number $t \in (0,1)$ where $\textrm{lim}_{n \to \infty} a_n t^n = 0$?
I think if $\lim_{n \to \infty} a_n t^n = 0$, then the limsup of $\sqrt[n]{a_n}$ is finite because: for all $\epsilon > 0$ there is $N$ such that $a_n t^n < \epsilon$ for $n > N$, so $\sqrt[n]{a_n} < \frac{\epsilon^{1/n}}{t}$, and taking limsup gives you $\textrm{limsup}_{n \to \infty} \sqrt[n]{a_n} < \frac{1}{t} < \infty$.
Suppose $\lim \sup\sqrt[n]{a_n}=\rho<\infty$. It means the power series $\sum a_n t^n$ converges for some $t>0$. More precisely, for sufficiently large $n$ we have $a_n<(\rho+\epsilon)^n$, so $t<1/(\rho+\epsilon)$ suffices to make the series converge. Hence $\lim a_n t^n=0$ for some $0<t<1$(one may take $t$ arbitrarily small).
Suppose $\lim a_n t^n=0$ for some $t>0$. It's easy to see that $\sum a_n\left(\frac{t}{2}\right)^n$ converges, hence $\lim \sup\sqrt[n]{a_n}\leq\frac{2}{t}<\infty$.