How to prove $\sum_{k=1}^n \frac{1}{2^k}$ is a Cauchy Sequence.

Question

I know that for a sequence to be a Cauchy sequence $\forall\epsilon > 0$ , $\exists M \in \mathbb R$ such that if $n,m > M$ then $|a_{n} - a_{m}| < \epsilon$

To start I have been told that $M = \frac{1}{\epsilon}$ however I am not sure how to proceed from here. I am mostly confused on the structure of the Cauchy sequence proofs. Do I proceed to manipulate the expression $|a_{n} - a_{m}|$ like this? $$\left\lvert\frac{1}{2^m} - \frac{1}{2^n}\right\rvert$$ If so, where do I go from here? Thank you.

Answer 1

Let $a_n=\sum_{k=1}^n\frac1{2^k}$. Then, we have for $n>m$

$$\begin{align}|a_n-a_m|&=\left|\sum_{k=m+1}^n\frac1{2^k}\right|\\\\ &=\frac1{2^m}-\frac1{2^n}\\\\ &=\frac1{2^m}\left(1-\frac{1}{2^{n-m}}\right)\\\\&\le \frac1{2^m}\\\\ &<\varepsilon \end{align}$$

whenever $n>m>\frac{\log(1/\varepsilon)}{\log(2)}$