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Math Help - Cauchy/convergent sequence

  1. #1
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    Cauchy/convergent sequence

    Question: Let {a_n} be a sequence of rational numbers defined as follows:

    a_1 = 1
    a_(n+1)=a_n + 1/(3^n)

    for all n > 1.

    1) Show that {a_n} is a Cauchy sequence and hence convergent
    2) Find it's limit.


    So.. I am a bit lost for this one. I tried using induction to bound a_n by 1 and 2, but I couldn't bound it above. I figured if I bound it above, and prove that it's a non-decreasing function, then I could prove that it converges, and thus that it is Cauchy...

    Any tips for me to get started on this? Thanks!
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  2. #2
    MHF Contributor chisigma's Avatar
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    Re: Cauchy/convergent sequence

    Quote Originally Posted by limddavid View Post
    Question: Let {a_n} be a sequence of rational numbers defined as follows:

    a_1 = 1
    a_(n+1)=a_n + 1/(3^n)

    for all n > 1.

    1) Show that {a_n} is a Cauchy sequence and hence convergent
    2) Find it's limit.


    So.. I am a bit lost for this one. I tried using induction to bound a_n by 1 and 2, but I couldn't bound it above. I figured if I bound it above, and prove that it's a non-decreasing function, then I could prove that it converges, and thus that it is Cauchy...

    Any tips for me to get started on this? Thanks!
    The difference equation can be written as...

    \Delta_{n}= a_{n+1}-a_{n}= \frac{1}{3^{n}}\ ,\ a_{1}=1 (1)

    ... and its solution is...

    a_{n}= a_{1} + \sum_{k=1}^{n-1} \Delta_{k}= a_{1} + \sum_{k=1}^{n-1} \frac{1}{3^{k}} (2)

    From (1) and (2) You derive that is...

    \lim_{n \rightarrow \infty} a_{n}= 1 + \sum_{k=1}^{\infty} \frac{1}{3^{k}}= 1 + \frac{1}{3\ (1-\frac{1}{3})}= 1+\frac{1}{2}= \frac{3}{2} (3)

    Kind regards

    \chi \sigma
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  3. #3
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    Re: Cauchy/convergent sequence

    riight! sum of geometric series. I'm not 1000% sure I can use this in my class right now, but it's worth a try. thank you!
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