Suppose $\lim_{n\to\infty} {\frac{a_{n+1}}{a_n}} = \frac{1}{2}$, prove $\lim_{n\to\infty}{a_n}=0$

You're on the right track, now WLOG assume $\epsilon < {1\over 2}$ so that after you reach the appropriate $N\in\Bbb N$ when your hypothesis starts to hold, with $0<a=\epsilon + {1\over 2} < 1$ and then you have $|a_{N+k}|<a^k|a_N|\to 0$ as $k\to\infty$ proving the result. You can even quantify it if you like, since

$$a^k|a_N|<\epsilon\iff k>{\log\epsilon -\log|a_N|\over\log a}$$

where the inequality switches direction because we divide by the negative number $\log a$.