Suppose $(X,d)$ is a metric space. Suppose we have constructed a sequence of open balls $(B(x_n,1/2^{n-1}))_{n\geq 1}$ in $X$ such that $x_n\in B(x_{n-1},1/2^{n-2})$ for all $n\geq 2$. How can I show that this sequence is Cauchy?
Attempt:
Note that for all $k\geq 1$, we have $d(x_k,x_{k-1})\leq 1/2^{k-1}$. We have to show that for all $\epsilon>0$, there exists $n\geq 1$ such that $$d(x_p,x_q)<\epsilon$$ for all $n\leq p,q$. Fix $\epsilon>0$. There exists $n\geq 1$ such that $1/2^{n-1}\leq1/n<\epsilon$. Pick $p,q\geq n$ and assume $p<q$. I have to show $$d(x_p,x_q)\leq\ldots<\epsilon.$$ What should I do next? How can I fill in the "$\ldots$"?
$$d(x_p, x_q) \le d(x_p, x_{p+1}) + d(x_{p+1}, x_{p+2}) + \cdots + d(x_{q-1}, x_q) \le \sum_{j=p-1}^{q-2} 2^{-j} \le \sum_{j=p-1}^\infty 2^{-j} = 2^{-p+2}$$