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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?

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  • $\begingroup$ @ first step you sure that converges? or want to proove Cauchy sequence without knowing convergence? $\endgroup$
    – Khosrotash
    Sep 28, 2019 at 18:42
  • $\begingroup$ @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 $\endgroup$
    – Algorisum
    Sep 28, 2019 at 18:45
  • $\begingroup$ 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. $\endgroup$
    – Lee Mosher
    Sep 28, 2019 at 20:25

2 Answers 2

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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 $m,n \to +\infty$

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  • $\begingroup$ Can you explain the last inequality? I dont see how that finite sum is less than $\frac{n}{3^{n+1}}$. $\endgroup$ Sep 28, 2019 at 20:05
  • $\begingroup$ @NicholasRoberts i edited my answer ...thanks for the correction.. $\endgroup$ Sep 28, 2019 at 20:11
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    $\begingroup$ 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. $\endgroup$ Sep 28, 2019 at 20:12
  • $\begingroup$ It is good now +1 $\endgroup$ Sep 28, 2019 at 20:14
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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 . $$|\frac{a_{n+1}}{3^{n+1}}+\frac{a_{n+2}}{3^{n+2}}+\frac{a_{n+3}}{3^{n+3}}+\cdots|+|\frac{a_{m+1}}{3^{m+1}}+\frac{a_{m+2}}{3^{m+2}}+\frac{a_{m+3}}{3^{m+3}}+\cdots|\leq \\ |\frac{1}{3^{n+1}}+\frac{1}{3^{n+2}}+\frac{1}{3^{n+3}}+\cdots|+|\frac{1}{3^{m+1}}+\frac{1}{3^{m+2}}+\frac{1}{3^{m+3}}+\cdots|$$

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