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Math Help - bolzano weirshtrass question

  1. #1
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    bolzano weirshtrass question

    the question in this link:

    http://img135.imageshack.us/img135/7662/60430495yx6.gif

    i only know that these series are bounded
    so i know that that there is at least one sub sequence which converges to
    one of the bounds of each sequence
    which gives me nothing because if it were converges
    then i could transform lim inf An ==> lim An
    because every sub sequence converges to the same limit which is the
    main limit of the sequence.

    what to do here?
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  2. #2
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    Let us prove: \liminf a_n \cdot \liminf b_n \leq \liminf a_nb_n.
    (If \{a_n\},\{b_n\} are bounded then so is \{a_nb_n\}).

    Let a = \liminf a_n, well this means a = \lim \bigg( \inf \{ a_k | k\geq n\} \bigg) .
    The sequence, \inf \{ a_k | k\geq n \} is non-increasing therefore the limit, a,
    is the greatest lower bound, so a\leq \inf \{ a_k | k\geq n\} for all n\geq 1.
    For n=1 we see a\leq \inf \{ a_1,a_2,... \} therefore a is a lower bound for \{a_1,a_2,...\}.
    This means, a\leq a_n for all n\geq 1.

    Likewise if b = \liminf b_n = \lim \bigg( \inf \{ a_k | k\geq n\} \bigg) then b\leq b_n for all n\geq 1.

    We are told that 0\leq a_n,b_n which means 0\leq a,b.
    Thus, we can write inequality, ab \leq a_n b_n for all n\geq 1.
    This means that, ab \leq \inf\{ a_1b_1,a_2b_2,... \} \leq \inf \{ a_kb_k | k\geq n \} for all n\geq 1.
    Thus, ab is a lower bound for the sequence \inf \{ a_kb_k | k\geq n\}.
    But this 'inferior sequence' is non-increasing so it has a limit.
    Thus, ab \leq \lim \bigg( \inf \{ a_kb_k| k\geq n\} \bigg) = \liminf (a_nb_n).
    This completes the proof.

    You can try doing the other problems now.
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  3. #3
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    regarding this definition

    <br /> <br />
a = \lim \bigg( \inf \{ a_k | k\geq n\} \bigg) <br />

    the sequence
    <br /> <br />
\inf \{ a_k | k\geq n \}<br />

    is non decreasing . its inf gets bigger or not changing in each following sequence
    so its limit is its least upper bound
    not like you said "greatest lower bound"

    am i correct??
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