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 $|x_{n+1} - x_n| < \large\frac{1}{2^n}$ $\forall n > N$, and I want to transition that to a statement of $|x_{m} - x_n| < \large\frac{1}{2^n}$ $\forall n > m$. Then the right side just becomes an epsilon as n goes infinity.
 A: Observe that by hypothesis, $$\sum_{k=n}^{\infty} |x_{n+1} - x_{n} | < \frac{1}{2^{n-1}}.$$ Therefore, you can show that $\{x_n\}$ is Cauchy, since given any $\epsilon > 0, \exists N$ so that $\frac{1}{2^{N-1}} < \epsilon$. Therefore, $\forall n,m > N, |x_n - x_m| \le \sum_{n=N}^{\infty} |x_{n+1} - x_n| < \epsilon$. There is a hidden triangle inequality in this last step, but I think you've got it from here.
A: Let $m,n\gt N$, with $m\lt n$. By the Triangle Inequality, we have
$$|x_n-x_m|\le |x_{m+1}-x_m| +|x_{m+2}-x_{m+1}| +\cdots +|x_n-x_{n-1}|.$$
The first term on the right is $\lt \frac{1}{2^m}$, the second term is $\lt \frac{1}{2^{m+1}}$, and so on. So the full sum on the right is $\lt \frac{1}{2^{m-1}}$.
For any given $\epsilon\gt 0$, we can find an $N_1\gt N$ such that $\frac{1}{2^{m-1}}\lt  \epsilon$. 
It follows that if  $N_1\lt m\lt n$, then $|a_m-a_n|\lt \epsilon$.
Remark: Informally, for large $m$, we have that  $x_{m+1}$ is very close to $x_m$, and $x_{m+2}$ is very close to $x_{m+1}$, and so on. So how far can $x_n$ be from $x_m$? The point is that even in the worst case, if the errors add up, $x_n$ must be close to $x_m$. 
Note that if in the problem $\frac{1}{2^n}$ is replaced by $\frac{1}{n}$, the argument breaks down. For the "tail" $\frac{1}{m}+\frac{1}{m+1}+cdots$ of the harmomic series is infinite. 
A: Let $m=n+1$.
Since $n>N$, then $m>N$. So it becomes $|x_m-x_n|<\epsilon$ (letting $\epsilon=\frac{1}{2^n}$) for $m,n>N$ which is the definition of Cauchy sequence.
