Show that a sequence of reals is cauchy So I right away notice that this $\large\frac{1}{2^n}$ will go to zero, and this can just be replaced by an epsilon.

However, it is not apparent to me how to transition about the statement given about the two sequential elements to any arbitrary element. (if that language is bad, I mean i'm given that $|x_{n+1} - x_n| &lt; \large\frac{1}{2^n}$ $\forall n &gt; N$, and I want to transition that to a statement of $|x_{m} - x_n| &lt; \large\frac{1}{2^n}$ $\forall n &gt; m$. Then the right side just becomes an epsilon as n goes infinity.
Observe that by hypothesis, $$\sum_{k=n}^{\infty} |x_{n+1} - x_{n} | &lt; \frac{1}{2^{n-1}}.$$ Therefore, you can show that $\{x_n\}$ is Cauchy, since given any $\epsilon &gt; 0, \exists N$ so that $\frac{1}{2^{N-1}} &lt; \epsilon$. Therefore, $\forall n,m &gt; N, |x_n - x_m| \le \sum_{n=N}^{\infty} |x_{n+1} - x_n| &lt; \epsilon$. There is a hidden triangle inequality in this last step, but I think you've got it from here.
Let $m=n+1$.
Since $n&gt;N$, then $m&gt;N$. So it becomes $|x_m-x_n|&lt;\epsilon$ (letting $\epsilon=\frac{1}{2^n}$) for $m,n&gt;N$ which is the definition of Cauchy sequence.