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. :)