Suppose that $\{x_n\}_n$ satisfies $|x_n - x_{n+1}|\leq\frac{1}{2^n},\;\forall n\in\mathbb{N}$. Show that $\{x_n\}$ converges.

Question

One hint I was given was show that $\{x_n\}_n$ is Cauchy. How can I do this?

Answer 1

Hint:

If you know that $|x_n-x_{n+1} | \leq 2^{-n}$ for every $n$, then
$$|x_n - x_{n+2}| \leq |x_n-x_{n+1}| + |x_{n+1}-x_{n+2}| \leq 2^{-n} + 2^{-(n+1)}$$

Now you can prove that $$\sum_{k=n}^{m} 2^{-k} = 2^{-m} (2^m-1) - 2^{-n}(2^n-1)= 2^{-n}-2^{-m}$$

Can you see how to proceed?