Proof That a Sequence is Cauchy $\{a_n\}$ is a sequence such that $|a_{n+1} - a_n| < 5^{-n}$ for all $n>0$.
How to prove that this sequence is a Cauchy sequence?
 A: Hint The above means that the sum $\displaystyle\sum_{n\geqslant 1}|a_{n}-a_{n+1}|$ converges. We usually say in this case the sequence $\{a_n\}$ is of bounded variation. You can change $d(x,y)=|x-y|$ by any other metric in any other metric space. 
Now given $\varepsilon >0$, $m>n$, write $|a_n-a_m|\leqslant |a_n-a_{n+1}|+|a_{n+1}-a_{n+2}|+\cdots+|a_{m-1}-a_{m}|$ and use the above to show this can be made $<\varepsilon$ for any $\varepsilon>0$ by taking $m,n$ larger than some $M$ depending on $\varepsilon$.
Spoiler 

Since $\displaystyle\sum_{n\geqslant 1}|a_{n}-a_{n+1}|$ converges, given $\varepsilon >0$; you can choose $M=M_\varepsilon $ such that $$\sum_{k=m}^n|a_{k+1}-a_k|<\varepsilon$$ whenever $m>n>M$. This gives that $|a_n-a_m|<\varepsilon$ by the triangle inequality, so the sequence is indeed Cauchy.

In particular, you're proving part of the following claim.

Claim Every sequence in a metric space $(M,\rho)$ that is of bounded variation is Cauchy. On the other hand, every Cauchy sequence contains a subsequence that is of bounded variation.

