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Math Help - HELP with cauchy sequences

  1. #1
    Junior Member mremwo's Avatar
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    HELP with cauchy sequences

    I am so confused with Cauchy Sequences. I was sick the day we learned about them. Here are two problems that I am absolutely unsure about. (just two of many)

    1. Prove straight from the definition that (n + \frac{(-1)^n}{n}) is NOT a Cauchy Sequences

    and

    2. Let (x_{n}) be a Cauchy Sequence such that x_{n} is an integer for every n \in  N. Show that (x_{n}) is ultimately constant.

    As I said, I am so lost. Please help? Thanks so much.
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  2. #2
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    Quote Originally Posted by mremwo View Post
    I am so confused with Cauchy Sequences. I was sick the day we learned about them. Here are two problems that I am absolutely unsure about. (just two of many)

    1. Prove straight from the definition that (n + \frac{(-1)^n}{n}) is NOT a Cauchy Sequences

    and

    2. Let (x_{n}) be a Cauchy Sequence such that x_{n} is an integer for every n \in  N. Show that (x_{n}) is ultimately constant.

    As I said, I am so lost. Please help? Thanks so much.

    And that is why you don't miss maths classes in college/university unless your city has

    been nuked or you're completely dead...honest!

    1) We get that for or any \displaystyle{n\in\mathbb{N}\,,\,\,\left|n+\frac{(-1)^n}{n}-\left((n+1)+\frac{(-1)^{n+1}}{n+1}\right)\right|=}

    \displaystyle{\left\{\begin{array}{ll}\left|-1-\frac{1}{n}-\frac{1}{n+1}\right|=\frac{n^2+3n+1}{n(n+1)}&\mbox  {, if n is odd}\\{}\\\left|-1+\frac{1}{n}+\frac{1}{n+1}\right|=\frac{n^2+n-1}{n(n+1)}}&\mbox{, if n is even}\end{array}\right.}

    It's easy to see that in both cases we get that the resulting fraction is greater than or

    equal 0.5, so choosing \epsilon=0.1, say, and calling the sequence a_n , we get that for any

    n, |a_n-a_{n+1}|\nless \epsilon , which contradicts the definition of Cauchy seq. (just take m = n+1)

    2) For such a sequence, if we have different elements, then |x_n-x_m|\geq 1 , but as the

    seq. is Cauchy take say \displaystyle{\epsilon=\frac{1}{2}\Longrightarrow \exists N\in\mathbb{N}\,\,s.t.\,\,\forall\,n,m>N\,,\,\,|x_  n-x_m|<\epsilon\Longleftrightarrow |x_n-x_m|=0...and end

    now the argument.

    Tonio
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by tonio View Post
    And that is why you don't miss maths classes in college/university unless your city has

    been nuked or you're completely dead...honest!
    Remember the Mathematicians code: "Them's that fall behind are left behind" Arrrgh..

    CB
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  4. #4
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    Can you use the fact that "a sequence of real numbers is Cauchy if and only if it converges"?

    If so, they you can show that (1) is NOT a Cauchty sequence by showing that it is not convergent. And you can do that by showing it is not bounded.


    For (2) show that |a_{n+1}- a_n|< \epsilon, where a_{n+1} and a_n are integers, requires that a_{n+1}=  a_n.
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