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A formal proof for an affirmation about sequences

  1. Nov 1, 2013 #1
    1. The problem statement, all variables and given/known data

    Prove that:

    The sequence [itex] x_n \to x [/itex] if and only if there is a [itex] M > 0 [/itex] such that [itex]\forall \epsilon > 0[/itex] , [itex]\exists n_\epsilon \in \mathbb{N} [/itex] and [itex]n\geq n_\epsilon [/itex] we have [itex] | x_n - x | < \epsilon M [/itex]


    2. Relevant equations

    The first implication "=>" is proved by choosing M = 1. Then, the problem statement is exactly the epsilon-convergence theorem.

    I don't know how to prove "<=".

    3. The attempt at a solution

    I tried to chose [itex] \epsilon = \epsilon_1 = \frac{\epsilon}{M}[/itex] , and I get that:

    [itex]\exists n_{\epsilon_1} \in \mathbb{N} [/itex] and [itex]n\geq n_{\epsilon_1} [/itex] we have [itex] | x_n - x | < \epsilon_1 M = \epsilon[/itex]

    So I found an [itex] n_{\epsilon_1} [/itex] such that ..... [itex] | x_n - x | < \epsilon[/itex]. But I should've found a rang that depends on epsilon, not on epsilon1. And this confuses me a little because I don't know for sure if I can state the following: I've found [itex] n_{\epsilon_1} [/itex] so, by substituing , I've found [itex] n_{\frac{\epsilon}{M}} [/itex], which depends on epsilon, but also depends on M, which is a constant, and the theorem is proved. I'm not convinced. Can anyone help me?

    P.S: I'm not a native speaker and I definitely have problems with this formal language used, so I hope you guys understood exactly what I said. To me, it seems clear. If it's not, please, let me know, so I can rephrase it. :D
     
  2. jcsd
  3. Nov 1, 2013 #2
    Stating this a little less formally, what you are really trying to show is that given an M the statement

    |##x_n## -x| < ##\epsilon##M (1)

    for any ##\epsilon## means |##x_n## -x| goes to zero as n gets large.

    So you want to show that if (1) is true then for any ##\epsilon## there is some ##N_{\epsilon}##such that for all n > ##N_{\epsilon}##
    |##x_n## -x| < ##\epsilon##.

    You set up an ##\epsilon_1 = \epsilon##/M which is correct. Now pick ##N_{\epsilon}## so that for all n > N |##x_n## -x| < ##\epsilon_1##M. Your hypothesis says this should be possible. Of course ##\epsilon_1##M = ##\epsilon##, so you have your result.

    What you said is either correct or very close to correct. You knew what you were trying to do. The problem is in stating is so that you fulfill your hypothesis and still get your result. I don't think it is easy even if you speak English well.
     
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