# 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?

• @ first step you sure that converges? or want to proove Cauchy sequence without knowing convergence? Sep 28, 2019 at 18:42
• @Khosrotash The problem told me to show the sequence converge by showing it is a Cauchy sequence. Hence I assume that the sequence converge to s Sep 28, 2019 at 18:45
• Circular proofs are invalid. If you wish to prove that the sequence converges, starting your proof by assuming that the sequence converges makes your proof circular and therefore invalid. Sep 28, 2019 at 20:25

## 2 Answers

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 $$m,n \to +\infty$$

• Can you explain the last inequality? I dont see how that finite sum is less than $\frac{n}{3^{n+1}}$. Sep 28, 2019 at 20:05
• @NicholasRoberts i edited my answer ...thanks for the correction.. Sep 28, 2019 at 20:11
• No problem. I believe you need to change the upper index of summation to m (it is currently n). Then that sum will go to zero as m and n go to infinity? I believe that is the way to conclude. Sep 28, 2019 at 20:12
• It is good now +1 Sep 28, 2019 at 20:14

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 . $$|\frac{a_{n+1}}{3^{n+1}}+\frac{a_{n+2}}{3^{n+2}}+\frac{a_{n+3}}{3^{n+3}}+\cdots|+|\frac{a_{m+1}}{3^{m+1}}+\frac{a_{m+2}}{3^{m+2}}+\frac{a_{m+3}}{3^{m+3}}+\cdots|\leq \\ |\frac{1}{3^{n+1}}+\frac{1}{3^{n+2}}+\frac{1}{3^{n+3}}+\cdots|+|\frac{1}{3^{m+1}}+\frac{1}{3^{m+2}}+\frac{1}{3^{m+3}}+\cdots|$$