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Let $s_n$ be a sequence such that $|s_{n+1} -s_n| < 2^{-n}$ for all $n\in \mathbb N$. Prove that $s_n$ is a Cauchy sequence and therefore a convergent sequence.

This is what I have so far. I'm assuming its similar to an infinite limit proof?

Proof: let $s_n$ be a sequence which satisfy the condition. We want to show for every $\epsilon >0$ there exists an $N>0$ such that if $n>N$, then $|s_{n+1} -sn|<2^{-n}$.

So we want:

\begin{equation} \begin{split} 2^{-n}<E &\Leftrightarrow 2^n>1/E \\ &\Leftrightarrow \ln (2^n)> \ln (1/E)\\ &\Leftrightarrow n \ln(2) > \ln 1-\ln E \\ &\Leftrightarrow n> -\frac{\ln E}{\ln 2} \end{split} \end{equation}

verification: assume $n >-\frac{ \ln E}{\ln 2}$. Then for every $E > 0$ there exists an $N>0$ such that if $n> N=-\frac{ \ln E}{\ln 2}$ then

$$|s_{n+1} -s_n|<2^{-n} <E$$

From this we know $|s_{n+1} -s_n|<2^{-n}<E$ and therefore a Cauchy sequence and also convergent.

Stephanie Hernandez
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    Where is $ s_{n} $? You should give more information, because a Cauchy sequence is not necessairily convergent, unless the space in which the sequence is defined is complete. Therefore you should specify if we are in a complete space. And write in latex – Joaquin Liniado May 11 '15 at 03:20
  • The question didn't specify. – Stephanie Hernandez May 11 '15 at 03:44

2 Answers2

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Hints/Suggestions:

You need to first understand the definition of Cauchy sequence. To prove $s_n$ is Cauchy you need to show that for every $\epsilon >0$, there exists a $N>0$ such that for all $\color{red}{m,n >N}$ we have $$|s_m-s_n| < \epsilon.$$ The inequality you are given $$|s_{n+1}-s_{n}| < \frac{1}{2^n}.$$ is for consecutive terms $n$ and $n+1$, whereas for Cauchy you need to show the first inequality holds for for all $\color{red}{m,n >N}$.

Think in terms of triangle inequality. Suppose $m=n+k$ for $k \geq 1$, then \begin{align*} |s_m-s_n| & =|s_{n+k}-s_n|\\ & =|s_{n+k}-s_{n+k-1}+s_{n+k-1}-s_{n+k-2}+s_{n+k-2} +\dotsb - s_{n}|\\ & \leq |s_{n+k}-s_{n+k-1}|+|s_{n+k-1}-s_{n+k-2}|+\dotsb + |s_{n+1}- s_{n}|\\ & \leq \frac{1}{2^{n+k}} + \frac{1}{2^{n+k-1}}+ \dotsb + \frac{1}{2^{n}}\\ & = \frac{1}{2^n} \left(\frac{1}{2^{k}} + \frac{1}{2^{k-1}}+ \dotsb + 1\right)\\ & = \frac{1}{2^n} \frac{\left(1-\frac{1}{2^{k+1}}\right)}{1-\frac{1}{2}}\\ & = \frac{1}{2^{n-1}}\left(1-\frac{1}{2^{k+1}}\right)\\ & \leq \frac{1}{2^{n-1}}. \end{align*} Now to get the very first inequality (with $\epsilon$) you want $$\frac{1}{2^{n-1}} < \epsilon.$$ See if you can proceed from here.

Anurag A
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For $$\epsilon > 0, m,n \geq N_0, |s_m-s_n| = |(s_m-s_{m-1})+(s_{m-1}-s_{m-2}) + \cdots + (s_{n+1}-s_n)|\leq |s_m-s_{m-1}|+|s_{m-1}-s_{m-2}|+\cdots +|s_{n+1}-s_n|\leq 2^{-m}+2^{m-1}+\cdots +2^{-n}=\dfrac{1}{2^n}\left(1+\dfrac{1}{2}+\cdots + \left(\dfrac{1}{2}\right)^{m-n}\right)= \dfrac{1}{2^{n-1}}\left(1-\dfrac{1}{2^{m-n+1}}\right)<\dfrac{1}{2^{n-1}}< \epsilon, n > 1+\log_{2}\left(\frac{1}{\epsilon}\right)=N_0$$

The conclusion follows....

DeepSea
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