proving convergence by showing a sequence is Cauchy

Question

Let $(x_n)_{ n=1}^{∞} $ be a sequence of real numbers such that $|x_{n+1} − x_n| ≤ \frac {1} {2^n}$ for all $n$.

Show that $(x_n) ^∞ _{n=1}$ converges.

It seems pretty clear that I can prove that this sequence is Cauchy, since the terms of the sequence are getting arbitrarily close to each other as $n$ gets large (since $\frac {1} {2^n}$ gets abritrarily small ). I'm thinking that $\epsilon$ is related to $\frac {1} {2^n}$, but I don't know where to start the rigorous proof. Any help is appreciated!

Answer 1

Hint: Let $m>n$. According to the triangular inequality:$$|x_m-x_n|\le\sum_{i=n}^{m-1}|x_{i+1}-x_i|.$$

Answer 2

Take $\varepsilon>0$ and take $N\in\mathbb N$ such that $\frac1{2^{N-1}}<\varepsilon$. Then, if $m>n\geqslant N$,\begin{align}\lvert x_m-x_n\rvert&\leqslant\lvert x_m-x_{m-1}\rvert+\lvert x_{m-1}-x_{m-2}\rvert+\cdots+\lvert x_{n+1}-x_n\rvert\\&\leqslant\sum_{k=n}^{m-1}\frac1{2^k}\\&<\sum_{k=n}^\infty\frac1{2^k}\\&=\frac1{2^{n-1}}\\&\leqslant\frac1{2^{N-1}}\\&<\varepsilon.\end{align}