Let $a_n$ be a sequence of positive real numbers satisfying $a_n$ < $\frac{1}{2}a_{n-1}$ for all $n \ge 2$. Prove that it converges to zero. Let $a_n$ be a sequence of positive real numbers satisfying $a_n$ < $\frac{1}{2}a_{n-1}$ for  all  $n \ge 2$. Use an appropriate theorem and the fact that $\lim_{n\to\infty} (\frac{1}{2})^n = 0$ to prove that $\lim_{n\to\infty} a_n = 0$.
My attempt: $a_n$ is a decreasing sequence because $a_n < \frac{1}{2}a_{n-1}$, for all $n \ge 2$. And we can continue: $\frac{1}{2}a_{n+1} < \frac{1}{2}a_{n}$; $\frac{1}{2}a_{n+1} < \frac{1}{2}a_{n} <\frac{1}{2}a_{n-1}$. Then, $(\frac{1}{2})^na_{n+1} < (\frac{1}{2})^na_{n} <(\frac{1}{2})^na_{n-1}$ is also true. I see that the squeeze theorem is applicable but I am not sure how to proceed from here. Please advise.
 A: Hint: $0 < a_n < a_1\cdot \left(\dfrac{1}{2}\right)^{n-1}, n \ge 2$ .
A: EDIT
One possible approach consists in noticing that:
\begin{align*}
0 < a_{n} = \frac{a_{n}}{a_{n-1}}\times\frac{a_{n-1}}{a_{n-2}}\times\ldots\frac{a_{2}}{a_{1}}\times a_{1} < a_{1}\times\left(\frac{1}{2}\right)^{n-1}
\end{align*}
Consequently, due to the arithmetic properties of numerical sequences, one gets the desired result:
\begin{align*}
0 < a_{n} < a_{1}\times \left(\frac{1}{2}\right)^{n-1} \Rightarrow 0\leq\lim_{n\to\infty}a_{n} \leq \lim_{n\to\infty}a_{1}\times\left(\frac{1}{2}\right)^{n-1} \Rightarrow \lim_{n\to\infty}a_{n} = 0.
\end{align*}
Hopefully this helps!
A: Only decreasing property is not enough, we should use here, that sequence is, also, bounded from below, for example, by $0$. Now we can say, that limit exists and write
$$0\leqslant L=\lim\limits_{n\to\infty}a_n\leqslant\lim\limits_{n\to\infty}\frac{1}{2}a_{n-1}=\frac{1}{2}L$$
this gives $L=0$, where we used $\lim\limits_{n\to\infty}a_n=\lim\limits_{n\to\infty}a_{n-1}$.
A: Testing
Note that for all $n\ge 1$ we have
\begin{align*}
a_n\le \frac{1}{2}a_{n-1}\le \dots \le \left(\frac{1}{2}\right)^{n-1}a_{1}
\end{align*}
Also note that for all $x>0$ exists $m\in \mathbb{N}$ such that  $\frac{1}{m}<x$, with this we can conclude that for all $\varepsilon>0$ exists $N\in \mathbb{N}$ such that
\begin{align*}
|a_n|\le \left(\frac{1}{2}\right)^{n-1}a_1<a_1\varepsilon
\end{align*}
for all $n\ge N$. In other words
\begin{align*}
\lim_{n \to \infty}a_n=0
\end{align*}
