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Math Help - [SOLVED] Convertence

  1. #1
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    [SOLVED] Convertence

    Let a_n be a sequence of real numbers such that |a_{n+1}-a_n|<\frac{1}{2^n} for n=1,2,\dots prove that the sequence converges.

    My first thought was to show that a_n was a Cauchy sequence but am having trouble doing so. Any ideas about what else I should try?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by putnam120 View Post
    Let a_n be a sequence of real numbers such that |a_{n+1}-a_n|<\frac{1}{2^n} for n=1,2,\dots prove that the sequence converges.

    My first thought was to show that a_n was a Cauchy sequence but am having trouble doing so. Any ideas about what else I should try?
    that is what you should try. here, i'll start you off.


    Note that |a_n - a_m| = |a_n - a_{n + 1} + a_{n + 1} - a_{n + 2} + a_{n + 2} - \cdots + a_{m - 1} - a_m|

    ........................... \le |a_n - a_{n + 1}| + |a_{n + 1} - a_{n + 2}| + \cdots + |a_{m - 1} - a_m| by the Triangle inequality


    can you continue?
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    Quote Originally Posted by maokid
    Let a_n be a sequence of real numbers such that |a_{n+1}-a_n|<\frac{1}{2^n} for n=1,2,\dots prove that the sequence converges.

    My first thought was to show that a_n was a Cauchy sequence but am having trouble doing so. Any ideas about what else I should try?
    In my Real Analysis course we had the following problem on the exam:
    Let a_n be a sequence so that:
    \lim \ |a_{n+1}-a_n| = \frac{1}{2}
    Prove that a_n is a convergent sequence.

    It is extremely similar to your problem. What Jhevon said is basically the solution, you should use Cauchy sequence.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    In my Real Analysis course we had the following problem on the exam:
    Let a_n be a sequence so that:
    \lim \ |a_{n+1}-a_n| = \frac{1}{2}
    Prove that a_n is a convergent sequence.

    It is extremely similar to your problem. What Jhevon said is basically the solution, you should use Cauchy sequence.
    yes, i remember that exam question, i liked the solution, for some reason, that's why i remembered it.


    @putnam120: another hint: note that you will need to use the sum formula for a geometric series here, so think about that if you get stuck
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    So from the hint I have that
    |a_n-a_m|\le |a_n-a_{n+1}+\dots -a_m|\le |a_n-a_{n+1}|+\dots + |a_{m-1}-a_m| < \frac{1}{2^n}+\dots +\frac{1}{2^{m-1}} < \frac{1}{2^{n-1}} for n>m.

    Is that right?
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    Quote Originally Posted by putnam120 View Post
    So from the hint I have that
    |a_n-a_m|\le |a_n-a_{n+1}+\dots -a_m|\le |a_n-a_{n+1}|+\dots + |a_{m-1}-a_m| < \frac{1}{2^n}+\dots +\frac{1}{2^{m-1}} < \frac{1}{2^{n-1}} for n>m.

    Is that right?
    No.

    |a_{n+1}-a_n| + ... + |a_{m+1}-a_m| \leq \frac{1}{2^n}+...+\frac{1}{2^m} = \frac{1}{2^m}\left( 1 + \frac{1}{2}+...+\frac{1}{2^{n-m}} \right).

    Now follow by the geometric sum.
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    ok i was just being lazy. but thanks 4 all the help.
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