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Thread: Prove sequence is Cauchy

  1. #1
    Senior Member Pinkk's Avatar
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    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 $\displaystyle |s_{n}-s_{m}|< \epsilon$.
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  2. #2
    Member mabruka's Avatar
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    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
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  3. #3
    Senior Member Pinkk's Avatar
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    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 $\displaystyle \epsilon > \frac{1}{2^{N}}$?
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  4. #4
    Member mabruka's Avatar
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    Yeah i am sorry i got distracted there! !

    Let me edit the mistake.
    Last edited by mabruka; Feb 17th 2010 at 04:59 PM.
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  5. #5
    Member mabruka's Avatar
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    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.
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  6. #6
    Senior Member Pinkk's Avatar
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    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?
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  7. #7
    Member mabruka's Avatar
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    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
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  8. #8
    Senior Member Pinkk's Avatar
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    Thanks!
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  9. #9
    Senior Member Pinkk's Avatar
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    One more quick question; would this be a Cauchy sequence if we replace $\displaystyle \frac{1}{2^{n}}$ with $\displaystyle \frac{1}{n}$?
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  10. #10
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Pinkk View Post
    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.
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  11. #11
    Senior Member Pinkk's Avatar
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    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.
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