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Math Help - Give Counterexample: b_(n+1)-b_n --> 0, then b_n --> L

  1. #1
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    Give Counterexample: b_(n+1)-b_n --> 0, then b_n --> L

    given: a_n = b_(n+1) - b_n

    [if the limit of a_n = 0, then (b_n) has a limit] = FALSE.

    please provide a counterexample. with all my heart i think the converse is true. if have already proven the converse of the converse.

    i have no idea where to even begin, since it seems obvious that ... should the difference between consecutive terms approach zero, then the terms approach an agreeable limit...

    i.e. i keep coming up with sequences that actually DO converge (e.g. (-1)^n/n ...)
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  2. #2
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    Quote Originally Posted by Skerven View Post
    given: a_n = b_(n+1) - b_n

    [if the limit of a_n = 0, then (b_n) has a limit] = FALSE.

    please provide a counterexample. with all my heart i think the converse is true. if have already proven the converse of the converse.

    i have no idea where to even begin, since it seems obvious that ... should the difference between consecutive terms approach zero, then the terms approach an agreeable limit...

    i.e. i keep coming up with sequences that actually DO converge (e.g. (-1)^n/n ...)
    if you're assuming that L is "finite", then a counter-example would be b_n = \ln n. the converse is obviously true.
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  3. #3
    Super Member redsoxfan325's Avatar
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    Another is b_n=H_n, where H_n=\sum_{k=1}^n\frac{1}{k}.
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  4. #4
    Senior Member bkarpuz's Avatar
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    Wink

    b(n):=\sqrt{n} for n\in\mathbb{N}_{0}.
    By the way, any function satisfying f,f^{\prime}>0>f^{\prime\prime} (these imply that a>0 tends to 0 asymptotically) with f(\infty)=\infty (this implies b diverges to \infty) can be your sequence b.
    Draw such a function on a paper and try to see what I mean...
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  5. #5
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    Quote Originally Posted by bkarpuz View Post
    By the way, any function satisfying f,f^{\prime}>0>f^{\prime\prime} (these imply that a>0 tends to 0 asymptotically) with f(\infty)=\infty (this implies b diverges to \infty) can be your sequence b.
    Yeah, as strange as it seems, I get it. f(x) -> Inf while f'(x) -> 0. Thanks for that generality.
    @redsoxfan325: I also thought of that odd series.
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