I found the next proof in my calculus notes:
Prove that any sequence $(x_n)\in\mathbb{R}$ that satisfies $|x_n-x_{n+1}|\leq \frac{1}{2^n}$, with $n\in\mathbb{N}$, converges.
The proof aims to conclude that the sequence is a Cauchy sequence, and therefore, it converges.
Proof: If $m>n,$ then $$|x_n-x_{n+1}|=|x_n-x_{n+1}+x_{n+1}-x_{n+2}+...+x_{m-1}-x_m|$$ $$\leq |x_n-x_{n+1}|+|x_{n+1}-x_{n+2}|+...+|x_{m-1}-x_m|$$ $$\leq\frac{1}{2^{n-1}}\sum_{k=1}^{m-n}\frac{1}{2^k}<\frac{1}{2^{n-1}}*1=\frac{1}{2^{n-1}}.$$Then $|x_n-x_m|<\frac{1}{2^{n-1}},$ for all $m>n\geq 1.$ Given $\varepsilon>0$, one can choose $N\in\mathbb{N}$ such that $n\geq N$ implies $\frac{1}{2^{n-1}}<\varepsilon$. Thus, if $m,n\geq N$ then $|x_n-x_m|<\varepsilon$ (Cauchy).
What I don't understand is how we can say that this $\sum_{k=1}^{m-n}\frac{1}{2^{n-1}}$ goes from $1$ to $m-n$
$$\begin{align}&\quad|x_n-x_{n+1}|+|x_{n+1}-x_{n+2}|+\cdots + |x_{m-1}-x_m|\\&=\sum_{k=n}^{m-1} |x_k-x_{k+1}|\\&\le\sum_{k=n}^{m-1} \frac1{2^k}\\&=\sum_{k=n}^{m-1}\frac 1{2^{n-1}}\frac1{2^{k-(n-1)}}\\&=\frac 1{2^{n-1}}\sum_{k=1}^{m-n}\frac1{2^k}\end{align}$$