Proof that $\frac{a_n}{3^n}$ is a Cauchy sequence that converges I'm trying to understand how to proof Cauchy sequence that converges. 
Given that let $a_i$ be a sequence of number such that $a_i \in \{-1, 0,1,\}$ for each i.
Let $s_n$ be a sequence that $s_n = \frac{a_1}{3} + \frac{a_2}{3^n} + \frac{a_3}{3^3} + ... + \frac{a_n}{3^n}$.
Proof:
Suppose $s_n$ converge to $s$, where $\lim s_n = s$. Let $\epsilon > 0$ then there exist $N \in \mathbb{N}$ such that 
$|s_n - s| < \frac{\epsilon}{2}$. Then for all $n, m \geq N$, we have 
$$|s_n - s_m| = |s_n - s + s - s_m| \leq |s_n - s| + |s - s_m|$$
$$= |\frac{a_n}{3^n} - s| + |\frac{a_m}{3^m} - s| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$$
Hence, we proved that $s_n$ is a Cauchy sequence. Therefore, the following sequence $s_n$ converges.
Can someone verify that if I did the proof correctly?
 A: If you went to show convergence by showing that the sequence is Cauchy then you cannot assume convergence a priori.(Since this is what you must show)
Let $m>n$ 
Then $$|s_m-s_n| \leq \sum_{k=n+1}^m\frac{|a_k|}{3^k} \leq \sum_{k=n+1}^m\frac{1}{3^{k}}=\frac{3}{2}\frac{1}{3^{n+1}}(1-\frac{1}{3^{m-n}}) \to 0$$ as $m,n \to +\infty$
A: Assume $n>m $ ,$s = \frac{a_1}{3} + \frac{a_2}{3^n} + \frac{a_3}{3^3} + ...\frac{a_m}{3^m}+... + \frac{a_n}{3^n}+\cdots$
$$|s_n - s_m| = |s_n - s + s - s_m| \leq |s_n - s| + |s - s_m|\\=
|(\frac{a_1}{3} + \frac{a_2}{3^n} + \frac{a_3}{3^3} + ...\frac{a_m}{3^m}+... + \frac{a_n}{3^n}) - s| + |s - (\frac{a_1}{3} + \frac{a_2}{3^n} + \frac{a_3}{3^3} + ...\frac{a_m}{3^m})|=\\
|\frac{a_{n+1}}{3^{n+1}}+\frac{a_{n+2}}{3^{n+2}}+\frac{a_{n+3}}{3^{n+3}}+\cdots|+|\frac{a_{m+1}}{3^{m+1}}+\frac{a_{m+2}}{3^{m+2}}+\frac{a_{m+3}}{3^{m+3}}+\cdots|$$ take over from here .
$$|\frac{a_{n+1}}{3^{n+1}}+\frac{a_{n+2}}{3^{n+2}}+\frac{a_{n+3}}{3^{n+3}}+\cdots|+|\frac{a_{m+1}}{3^{m+1}}+\frac{a_{m+2}}{3^{m+2}}+\frac{a_{m+3}}{3^{m+3}}+\cdots|\leq \\
|\frac{1}{3^{n+1}}+\frac{1}{3^{n+2}}+\frac{1}{3^{n+3}}+\cdots|+|\frac{1}{3^{m+1}}+\frac{1}{3^{m+2}}+\frac{1}{3^{m+3}}+\cdots|$$
