$(a_n)$ is a cauchy sequence, show that there exists a sequence $(a_{n_k}) \subseteq (a_n)$ such that $|a_{n_{k+1}} - a_{n_k} | < \frac{1}{2^k}$

Question

let $(a_n)$ be a cauchy sequence, show that there exists a sequence such that $(ax_{n_k}) \subseteq (a_n)$ such that $|a_{n_{k+1}} - a_{n_k} | < \frac{1}{2^k}$ for all $k$

since $a_n$ is cauchy we know that it converges,bounded and according to bolzano-weierstrass theorem we know that there exists a convergent sub sequence.

and according to cauchy definition a sequence $a_n$ is cauchy for any positive real number $\varepsilon$ there exists $N \in \Bbb N$ such that for all $n,m>N$ $|a_n-a_m| < \varepsilon$

I think that I need to construct a subsequence $a_{n_k}$ till I get to $|a_{n_{k+1}} - a_{n_k} | < \frac{1}{2^k}$ but I am having a hard time actually applying it

I thought about choosing an $\varepsilon$ each time for example if $\varepsilon = \frac{1}{2}$ that for $N_1$ we get $ |a_m-a_n| < \frac{1}{2}$ but I am not sure how to actually go on with this and how to conduct the proof

Thanks for any tips and help