# Prove sequence is Cauchy

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• Feb 17th 2010, 04:22 PM
Pinkk
Prove sequence is Cauchy
Let $\displaystyle (s_{n})$ be a sequence such that $\displaystyle |s_{n+1} - s_{n}| < 2^{-n}$ for all $\displaystyle n \in \mathbb{N}$. Prove that $\displaystyle (s_{n})$ is a Cauchy sequence.

How would I exactly go about this? My professor really didn't go over this. All she defined was that a Cauchy sequence is such that given $\displaystyle \epsilon > 0$ there exists $\displaystyle N$ such that if $\displaystyle n,m > N$, then $\displaystyle |s_{n}-s_{m}|< \epsilon$.
• Feb 17th 2010, 04:37 PM
mabruka
Given $\displaystyle \epsilon$ let $\displaystyle N \in \mathbb{N}$ be such that $\displaystyle \frac{1}{2^N}<\epsilon$.

Then for every $\displaystyle n,m \geq N$ (supose without loss of generality that m<n) we have that

$\displaystyle |s_m - s_n | \leq |s_m -s _{m+1} | + |s_{m+1} - s_{m+2} |+ \ldots +|s_{n-1} - s_n |$

$\displaystyle \leq \frac{n-m}{2^m} < \epsilon$

what do you think?

Errrr WAIT something is not right with the last inequality
• Feb 17th 2010, 04:42 PM
Pinkk
I understand the inequality up until the point where you arrive at $\displaystyle \frac{n-m}{2^{m}}$. How does the left side of the inequality lead to that, and how does that imply it is less than $\displaystyle \epsilon$ if we let $\displaystyle \epsilon > \frac{1}{2^{N}}$?
• Feb 17th 2010, 04:46 PM
mabruka
Yeah i am sorry i got distracted there!(Happy) !

Let me edit the mistake.
• Feb 17th 2010, 05:00 PM
mabruka
About how we got to $\displaystyle \frac{n-m}{2^{m}}$ we aplied the hypothesis for each term. How many terms we had? n-m terms.

What i mean is that every term has the form

$\displaystyle |s_{m+a} - s_{m+a+1}| <\frac{1}{2^{m+a}}<\frac{1}{2^m}$

because m<m+a

About hte other issue, i am thinking maybe we need another bound there because i cant see how to continue form here.
• Feb 17th 2010, 05:14 PM
Pinkk
Hmm I understand that there are $\displaystyle n-m$ terms, but I'm still not following how that inequality holds. For instance, if $\displaystyle n = m + 2$, we'd have $\displaystyle |s_{m}-s_{n}| \le |s_{m} - s_{m+1}| + |s_{m+1} - s_{n}| < \frac{1}{2^{m}} + \frac{1}{2^{n-1}}$

So we have that whole addition of those absolute values must be less than $\displaystyle \frac{1}{2^{m}} + \frac{1}{2^{m + 1}} + ... + \frac{1}{2^{n-1}}$. But how is THAT less than $\displaystyle \frac{n-m}{2^{m}}$?

Edit: Nevermind, I see how each term must be less than or equal to $\displaystyle \frac{1}{2^{m}}$ which is added together $\displaystyle n-m$ times.

So I understand that, but yes, how does that lead to the inequality involving the epsilon?
• Feb 17th 2010, 05:23 PM
mabruka
Okey , first choose N such that $\displaystyle \frac{1}{2^N}<\epsilon$. Then for $\displaystyle n,m \geq N$ , $\displaystyle n<m$

$\displaystyle |s_m - s_n | \leq |s_n -s _{n+1} | + |s_{n+1} - s_{n+2} |+ \ldots +|s_{m-1} - s_m |$

$\displaystyle <\frac{1}{2^n}+\frac{1}{2^{n+1}}+\ldots +\frac{1}{2^{m-1}}$

$\displaystyle = \frac{1}{2^n} \left(1+\frac{1}{2}+\ldots+\frac{1}{2^{m-(n+1)}} \right) <\frac{1}{2^{n-1}}<\frac{1}{2^N}<\epsilon$

I think it is correct now :)
• Feb 17th 2010, 05:33 PM
Pinkk
Thanks!
• Feb 17th 2010, 05:49 PM
Pinkk
One more quick question; would this be a Cauchy sequence if we replace $\displaystyle \frac{1}{2^{n}}$ with $\displaystyle \frac{1}{n}$?
• Feb 17th 2010, 07:30 PM
Drexel28
Quote:

Originally Posted by Pinkk
One more quick question; would this be a Cauchy sequence if we replace $\displaystyle \frac{1}{2^{n}}$ with $\displaystyle \frac{1}{n}$?

What do you think? The way the question is phrased should give the answer away.
• Feb 17th 2010, 07:36 PM
Pinkk
I don't think so, but I am not sure how to actually show it (at least only based on the knowledge we have learned so far in class). Because then we'd have (if we assume WLOG that $\displaystyle m>n$, $\displaystyle |s_{n} - s_{m}| < \frac{1}{n} + \frac{1}{n+1} + ... + \frac{1}{m-1}$, and in order for that to be Cauchy, that whole inequality must also be less than any given $\displaystyle \epsilon > 0$ where $\displaystyle N$ exists and $\displaystyle n,m > N$. But if that were true for ALL $\displaystyle n,m$, then $\displaystyle \frac{1}{n} + \frac{1}{n+1} + ... + \frac{1}{m-1}$ would have to be less than some number for ALL n,m, which doesn't look like the case...ugh, I have a vague idea but I have no idea how to formulate into coherent/logical/proof-worthy words. (Headbang)(Headbang)