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.

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!
