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Math Help - Convergence

  1. #1
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    Convergence

    Suppose 0<r<1 and |a(sub n+1)-a(sub n)|<r^n for all n elements of the natural numbers. Prove that {a(sub n)}n=1 to infinity converges. Here is my work:

    Since 0<r<1, then r^n --> 0. Since |a(sub n+1)-a(sub n)|<r^n for all n elements of the natural numbers, then this implies that |b(sub n+1)-a(sub n)| --> 0. Therefore, a(sub n) is Cauchy, and therefore must converge.

    I feel like all I have done is repeat what was given in the problem. Is there something else I need to add or clarify?
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  2. #2
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    Re: Convergence

    You haven't specified what b_{n+1} is. But you're on the right track - you can get a simple proof that works by showing that it's a Cauchy sequence.
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  3. #3
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    Re: Convergence

    Since r^n --> 0, then |a(sub n+1)-a(sub n)|<0 for all n elements of the natural numbers. Let E>0. Since a(sub n) --> A, there exists n in the natural numbers such that |a(sub n)-A|<E/2 whenever n>=N. If m,n>=N, then |a(sub n)-a(sub m)|<=|a(sub n)-A|+|a(sub m)-A|<E/2+E/2=E.

    Does this work?
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  4. #4
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    Re: Convergence

    Quote Originally Posted by lovesmath View Post
    Since r^n --> 0, then |a(sub n+1)-a(sub n)|<0 for all n elements of the natural numbers. Let E>0. Since a(sub n) --> A...
    Does this work?
    That won't work to prove convergence, since you're assuming convergence.

    Here's a hint for setting it up to show that it's a Cauchy sequence:

    |a_{n+k} - a_{n}| = |(a_{n+k} - a_{n+k-1}) + (a_{n+k-1} - a_{n+k-2}) + ... + (a_{n+1} - a_{n})|

    \le |a_{n+k} - a_{n+k-1}| + |a_{n+k-1} - a_{n+k-2}| + ... + |a_{n+1} - a_{n}|.
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  5. #5
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    Re: Convergence

    Quote Originally Posted by lovesmath View Post
    Since r^n --> 0, then |a(sub n+1)-a(sub n)|<0 for all n elements of the natural numbers. Let E>0. Since a(sub n) --> A, there exists n in the natural numbers such that |a(sub n)-A|<E/2 whenever n>=N. If m,n>=N, then |a(sub n)-a(sub m)|<=|a(sub n)-A|+|a(sub m)-A|<E/2+E/2=E.
    @lovesmath why don't you learn to make your posts readable?
    LaTeX is not that difficult.
    [tex] |a_{n+1}-a_n|<r^n [/tex] gives  |a_{n+1}-a_n|<r^n . Is that nice?

    Can you show that \left| {a_{n + 1} } \right| \leqslant \sum\limits_{k = 1}^n {r^k }~?
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