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

My start on this was to state the limit in terms of the definition and work on it from there:
$\displaystyle\lim_{n\to\infty} {\frac{a_{n+1}}{a_n}} = \frac{1}{2}$
$\implies \exists n_0 $ s.t., $\forall \epsilon>0, |\frac{a_{n+1}}{an} -\frac{1}{2}| < \epsilon$,   $(\forall n_0 > n)$
Using the triangle inequality I restated the inequality above as follows:
$|\frac{a_{n+1}}{an}| \leq |\frac{a_{n+1}}{an} -\frac{1}{2}| + \frac{1}{2} < \epsilon$ 
$\implies  |\frac{a_{n+1}}{an}| < \epsilon + \frac{1}{2}$
$\implies  |{a_{n+1}}| < |a_n|(\epsilon + \frac{1}{2})$
At this point I am a bit of a loss. Am I permitted to generalize the above result as:
$\implies  |{a_{n+k}}| < |a_n|(\epsilon + \frac{1}{2})^k$  ?
And if so, is there some way I can utilize this fact to arrive at a proof of $a_n \rightarrow 0$ ?
Thank you in advance for any help. :)
 A: 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 $\log a$.
A: You can find some n so that all the ratios are under 3/4 (e = 1/4) so the limit is less than  (a sub n) × (3/4) ^ m as m goes to infinity.
A: More generally, if each $a_n\ne  0$ and $1>V=\lim_{n\to \infty}\sup_{m\geq n}|\frac {a_{m+1}}{a_m}|$ then $\lim_{n\to \infty}a_n=0.$
Proof:  Let $n_0$ be such that $\forall n\geq n_0\left(\; |\frac {a_{n+1}}{a_n}|\leq \frac {1+V}{2}\;\right).$  Now for $n\geq n_0$ we have $$|a_{n+1}|=|a_{n_0}|\cdot \prod_{j=n_0}^n\left|\frac {a_{j+1}}{a_j}\right|\leq |a_{n_0}|\left(\frac {1+V}{2}\right)^{n-n_0+1}.$$ The far RHS  expression goes to $0$ as $n\to \infty$ because $0<\frac {1+V}{2}<1.$
