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Math Help - Limits

  1. #1
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    Limits

    I really need help with this one. Is there anyone who can get me started? thanks
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    Quote Originally Posted by Sasbe View Post
    I really need help with this one. Is there anyone who can get me started? thanks
    Well, it's been a few weeks since I've done limits, but if n goes to infinite, then you're basically having every n variable go off to 0, so if you substitute 0 in, you're left with lim a(0)+1-a(0)/0=0 and 1/0=0 so that should be your proof.
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    Quote Originally Posted by emttim84 View Post
    Well, it's been a few weeks since I've done limits, but if n goes to infinite, then you're basically having every n variable go off to 0, so if you substitute 0 in, you're left with lim a(0)+1-a(0)/0=0 and 1/0=0 so that should be your proof.
    I don't get it. It's not possible to divide by 0, right?
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  4. #4
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    Quote Originally Posted by Sasbe View Post
    I really need help with this one. Is there anyone who can get me started? thanks
    i'll try to prove it like this.. since \lim_{n\rightarrow \infty } \, a_{n+2} - a_n = 0, then a_{n+2} - a_n is bounded, say by M

     \implies |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 (why?)  \implies -M < a_{n+1} - a_n < M

    so when we multiply  \frac{1}{n} to all, we have \frac{-M}{n} < \frac{a_{n+1} - a_n}{n} < \frac{M}{n} and if  n \rightarrow \infty, we have, 0 < \frac{a_{n+2} - a_n}{n} < 0 and by squeeze theorem, we get  \lim_{n \rightarrow \infty} \frac{a_{n+1} - a_n}{n} = 0 . QED
    Last edited by kalagota; November 26th 2007 at 05:55 AM.
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    Quote Originally Posted by kalagota View Post
    i'll try to prove it like this.. since \lim_{n\rightarrow \infty } \, a_{n+2} - a_n = 0, then a_{n+2} - a_n is bounded, say by M

     \implies |a_{n+2} - a_n| \leq M \implies -M \leq a_{n+2} - a_n \leq M

    so when we multiply  \frac{1}{n} to all, we have \frac{-M}{n} \leq \frac{a_{n+2} - a_n}{n} \leq \frac{M}{n} and if  n \rightarrow \infty, we have, 0 \leq \frac{a_{n+2} - a_n}{n} \leq 0 and by squeeze theorem, we get  \lim_{n \rightarrow \infty} \frac{a_{n+2} - a_n}{n} = 0 . QED
    Thats right, but did you notice that it was suppose to be
     \lim_{n \rightarrow \infty} \frac{a_{n+1} - a_n}{n} = 0
    instead of
     \lim_{n \rightarrow \infty} \frac{a_{n+2} - a_n}{n} = 0
    ?

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  6. #6
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    Quote Originally Posted by Sasbe View Post
    Thats right, but did you notice that it was suppose to be
     \lim_{n \rightarrow \infty} \frac{a_{n+1} - a_n}{n} = 0
    instead of
     \lim_{n \rightarrow \infty} \frac{a_{n+2} - a_n}{n} = 0
    ?

    In that case you can write,
    a_{n+2}-a_n = (a_{n+2}-a_{n+1})+(a_{n+1}-a_n) since \lim a_{n+1}-a_n = \lim a_{n+2}-a_{n+1}=0 apply kologata's argument.
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    Quote Originally Posted by ThePerfectHacker View Post
    In that case you can write,
    a_{n+2}-a_n = (a_{n+2}-a_{n+1})+(a_{n+1}-a_n) since \lim a_{n+1}-a_n = \lim a_{n+2}-a_{n+1}=0 apply kologata's argument.
    It may indeed matter. Take the example a_n  = \left( { - 1} \right)^n  + \frac{1}{n}.
    It is clear that both n & n+2 are together even or odd.
    So \left( {a_{n + 2}  - a_n } \right) \to 0.

    However, \left( {a_{n + 1}  - a_n } \right) does not converge to zero.
    If n is even we get \left( {a_{n + 1}  - a_n } \right) = \left[ { - 1 + \frac{1}{{n + 1}}} \right] - \left[ {1 + \frac{1}{n}} \right] =  - 2 + \frac{1}{{n + 1}} - \frac{1}{n}.
    Whereas if n is odd we get \left( {a_{n + 1}  - a_n } \right) = \left[ {1 + \frac{1}{{n + 1}}} \right] - \left[ { - 1 + \frac{1}{n}} \right] = 2 + \frac{1}{{n + 1}} - \frac{1}{n}.

    The key here is being bounded. That is what needs to be shown for \left( {a_n } \right)
    Last edited by Plato; November 25th 2007 at 01:50 PM.
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  8. #8
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by Sasbe View Post
    Thats right, but did you notice that it was suppose to be
     \lim_{n \rightarrow \infty} \frac{a_{n+1} - a_n}{n} = 0
    instead of
     \lim_{n \rightarrow \infty} \frac{a_{n+2} - a_n}{n} = 0
    ?

    oh yeah, didn't notice it.. tnx

    i've edited the solution, try to check it..
    Last edited by kalagota; November 25th 2007 at 09:14 PM.
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    Quote Originally Posted by kalagota View Post
    i'll try to prove it like this.. since \lim_{n\rightarrow \infty } \, a_{n+2} - a_n = 0, then a_{n+2} - a_n is bounded, say by M

     \implies |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 (why?)  \implies -M < a_{n+1} - a_n < M

    so when we multiply  \frac{1}{n} to all, we have \frac{-M}{n} \leq \frac{a_{n+1} - a_n}{n} \leq \frac{M}{n} and if  n \rightarrow \infty, we have, 0 \leq \frac{a_{n+2} - a_n}{n} \leq 0 and by squeeze theorem, we get  \lim_{n \rightarrow \infty} \frac{a_{n+1} - a_n}{n} = 0 . QED
    How do we know this?
     |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

    I would understand this if |a_{n+2} - a_{n+1}| + |a_{n+1} - a_n| was less (not greater) than |a_{n+2} - a_{n+1} + a_{n+1} - a_n|, but its not.
    Last edited by Sasbe; November 26th 2007 at 01:24 AM.
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  10. #10
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    Quote Originally Posted by kalagota View Post
    i'll try to prove it like this.. since \lim_{n\rightarrow \infty } \, a_{n+2} - a_n = 0, then a_{n+2} - a_n is bounded, say by M
     \implies |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
    That still does not work!
    As I said above, you must show (if you can) that \left( {a_n } \right) is bounded.
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  11. #11
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by Plato View Post
    That still does not work!
    As I said above, you must show (if you can) that \left( {a_n } \right) is bounded.
    it is bounded.. take note that a convergent sequence is bounded.. so if you consider a_{n+2} - a_n as a single sequence (convergent to 0), then you can conclude that it is bounded!
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  12. #12
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by Sasbe View Post
    How do we know this?
     |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

    I would understand this if |a_{n+2} - a_{n+1}| + |a_{n+1} - a_n| was less (not greater) than |a_{n+2} - a_{n+1} + a_{n+1} - a_n|, but its not.
    yes, it could be equal.. but not on the last parts that i've already edited..
    try to group the terms: |(a_{n+2} - a_{n+1}) + (a_{n+1} - a_n)|, then use triangle inequality...
    Last edited by kalagota; November 26th 2007 at 05:58 AM.
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  13. #13
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    Quote Originally Posted by kalagota View Post
    it is bounded.. take note that a convergent sequence is bounded.. so if you consider a_{n+2} - a_n as a single sequence (convergent to 0), then you can conclude that it is bounded!

    The whole point is that \left( {a_n } \right) may not be a covergent sequence.

    Take a close look at the example that I gave above.
    In that example the \left( {a_n } \right) sequence does not converge. Now it is bounded.

    So from the given, you may not conclude that the sequence converges because there is a counterexample.
    How do you prove that it is bounded?
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  14. #14
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by Plato View Post
    The whole point is that \left( {a_n } \right) may not be a covergent sequence.
    Take a close look at the example that I gave above.
    In that example the \left( {a_n } \right) sequence does not converge. Now it is bounded.

    So from the given, you may not conclude that the sequence converges because there is a counterexample.
    How do you prove that it is bounded?
    Theorem: A convergent sequence is bounded.. (do i need to prove this?)

    the given was  \lim_{n \rightarrow \infty} a_{n+2} - a_n and not {a_n } alone..
    you should not put your interest with a_n since what we want is the sequence a_{n+1} - a_n and that sequence is different from a_n!

    if you find any "flaws" with my "proof" while i had used only the given and the definition and some theorems, point it out.. well, i would try to consult this with my professor tomorrow..
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    Quote Originally Posted by kalagota View Post
    if you find any "flaws" with my "proof" while i had used only the given and the definition and some theorems, point it out..
    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.
    Quote Originally Posted by Sasbe View Post
    How do we know this?
     |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
    I would understand this if |a_{n+2} - a_{n+1}| + |a_{n+1} - a_n| was less (not greater) than |a_{n+2} - a_{n+1} + a_{n+1} - a_n|, but its not.
    Don't you see? Sasbe had already pointed that out to you.
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