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?
Best Answer
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 $m,n \to +\infty$