<s>[INST] 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.
 [/INST] Hint: $0 < a_n < a_1\cdot \left(\dfrac{1}{2}\right)^{n-1}, n \ge 2$ .
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