Show that $a_{n+1} =a_n + \frac{1}{3^n}$ defines a Cauchy sequence I just need help. 
Let ${a_n}$ be the sequence defined by the relations $a_1 = 1$ and $a_{n+1} =a_n + \frac{1}{3^n}$ for $n = 1, 2,....$ Prove that ${a_n}$ is a Cauchy sequence using definition.
 A: It's easy to see that
$$a_n = \sum_{k=0}^{n}\frac{1}{3^k}$$
The RHS is the $n$'th partial sum of a convergent geometric series (with limit $3/2$), so $a_n$ is a convergent sequence, hence a Cauchy sequence.

Edit after the OP added "using definition":

Let $m > n$. Then
$$\begin{aligned}
a_m - a_n &= \sum_{k=n+1}^{m}\frac{1}{3^k} \\
&= \frac{1}{3^{n+1}}\sum_{k=0}^{m-n-1}\frac{1}{3^k} \\
&= \frac{1}{3^{n+1}}\frac{1 - 1/3^{m-n}}{1 - 1/3} \\
&= \frac{3}{2}\frac{1}{3^{n+1}}\left(1 - \frac{1}{3^{m-n}}\right) \\
\end{aligned}$$
Observe that
$$0 \leq 1 - \frac{1}{3^{m-n}} \leq 1$$
Therefore,
$$0 \leq a_m - a_n \leq \frac{3}{2} \frac{1}{3^{n+1}} = \frac{1}{2}\frac{1}{3^n}$$
Given $\epsilon > 0$, the RHS will be less than $\epsilon$ whenever
$$3^n > \frac{1}{2\epsilon}$$
A: A sequence is Cauchy if for any $\epsilon>0$, there exists $N$ such that for any $n,m>N$, we have $|a_n-a_m|<\epsilon$.
In your case, just observe that $|a_{n+1}-a_n|=\frac{1}{3n}$ for any $n \in \mathbb{N}$. Now, use this fact and triangle inequality, and you will prove your claim.
