Proof Verification- proving that a sequence is a Cauchy sequence

Question

Prove that \begin{array}{l}\left\{\frac{2^n-1}{2^n}\right\},\ n\in\mathbb{N}\\ \end{array} is a Cauchy sequence. This is what I have so far: Let $\epsilon>0$ and $m,n >k$. Then,$$\begin{gather}\frac{1}{2^m}+\frac{1}{2^n}<\frac{1}{2^k}+\frac{1}{2^k}\\ \therefore \frac{1}{2^m}+\frac{1}{2^n}<\frac{2}{2^k}\end{gather}$$ Then I was thinking of assuming $\frac{2}{2^k}<\epsilon$. Am I on the right track at all?

Answer 1

Why not just observe that $$\lim_n \frac{2^n-1}{2^n}= \lim_n (1-2^{-n}) = 1$$ and convergent sequences are Cauchy sequences?