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Math Help - Limits

  1. #16
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by Plato View Post
    OK. Here is the major flaw.
     \color{red} |a_{n+2} - a_{n+1} + a_{n+1} - a_n| \leq |a_{n+2} - a_{n+1}| + |a_{n+1} - a_n| \leq M \implies |a_{n+1} - a_n| < M
    There is absolutely no way that you can claim that. It is a false claim.
    i can.. take note that  |a_{n+2} - a_{n+1}| is always positive, therefore if i remove it, the remaining would be strictly less than M..
    where is the flaw there? you are the only person i know who disagreed with that.. as far as i'm concerned..


    Quote Originally Posted by Plato View Post
    Don't you see? Sasbe had already pointed that out to you.
    i have explained the reason why that is true..
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  2. #17
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    Quote Originally Posted by kalagota View Post
    i can.. take note that  |a_{n+2} - a_{n+1}| is always positive, therefore if i remove it, the remaining would be strictly less than M..
    where is the flaw there? you are the only person i know who disagreed with that.. as far as i'm concerned..




    i have explained the reason why that is true..
    Thats not the problem. IF this was true
     \color{red} |a_{n+2} - a_{n+1}| + |a_{n+1} - a_n| \leq M
    we could claim this
     \color{red} \implies |a_{n+1} - a_n| < M
    as you said,
    BUT here's the problem:we just know that
     \color{red} |a_{n+2} - a_{n+1} + a_{n+1} - a_n| \leq M \
    so we can't claim that something greater than  \color{red} |a_{n+2} - a_{n+1} + a_{n+1} - a_n|\ is less than M, because we don't know that.
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  3. #18
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    Opalg's Avatar
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    Quote Originally Posted by Sasbe View Post
    [See attached image below.]
    I think you need some fairly heavy-duty analysis to deal with this.

    Let ε > 0. since \lim_{n\to\infty}(a_{n+2}-a_n) = 0, there is an integer N such that |a_{n+2}-a_n|<\epsilon whenever n≥N. For any positive integer r, a_{N+2r} - a_N = (a_{N+2} - a_N) + (a_{N+4} - a_{N+2}) + (a_{N+6} - a_{N+4}) + \ldots + (a_{N+2r} - a_{N+2r-2}).

    Therefore |a_{N+2r} - a_N| \leqslant |a_{N+2} - a_N| + |a_{N+4} - a_{N+2}| + \ldots + |a_{N+2r} - a_{N+2r-2}| < r\epsilon. It follows that \frac{|a_{N+2r} - a_N|}{N+2r} < \frac{r\epsilon}{N+2r}\to \epsilon/2 as r→∞. Hence \frac{|a_{N+2r} - a_N|}{N+2r} < \epsilon for all sufficiently large r. In fact, since \frac{a_N}{N+2r}\to0 as r→∞, we even have \frac{|a_{N+2r}|}{N+2r} < \epsilon for all sufficiently large r. Putting n=N+2r (or n=(N+1)+2r, which works equally well), you see that \frac{|a_n|}n < \epsilon for all sufficiently large n.

    Finally, \frac{|a_{n+1} - a_n|}n \leqslant \frac{|a_{n+1}|}n + \frac{|a_n|}n < 2\epsilon for all sufficiently large n. Since ε was arbitrary, this shows that \frac{|a_{n+1} - a_n|}n \to 0 as n→∞.
    Attached Thumbnails Attached Thumbnails Limits-4521d1195933973-limits-mathprob.jpg  
    Last edited by Opalg; November 27th 2007 at 03:53 AM.
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  4. #19
    MHF Contributor kalagota's Avatar
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    after re-reading everything, i now realize my big mistake..

    thanks to you plato..
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