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Math Help - bounded sequence proof

  1. #1
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    bounded sequence proof

    A) Suppose that Sn = 0, if (Tn) is a bounded sequence, prove that lim (SnTn) = 0
    B) Show by an example that the boundeness of (Tn) is a necessary condition in part (A)
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  2. #2
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    Quote Originally Posted by luckyc1423 View Post
    A) Suppose that Sn = 0, if (Tn) is a bounded sequence, prove that lim (SnTn) = 0
    We need to show,

    |s_n*t_n|<e

    Note that |t_n|<=M (where M can be chosen to be non-zero).

    Then,

    |s_n*t_n|<=M*|s_n|<e thus, we need that, |s_n|<e/M

    Which is possible for some n>N because lim s_n = 0.
    Q.E.D.
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  3. #3
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    Quote Originally Posted by luckyc1423 View Post
    B) Show by an example that the boundeness of (Tn) is a necessary condition in part
    Theorem A convergent sequence is bounded.

    We have shown that,

    lim (s_n*t_n)=0

    Thus,

    |s_n*t_n|<= M

    By the previous theorem.

    BUT! {s_n} is also bounded by some non-zero konstant K because it is convergent.

    Thus,

    |s_n*t_n|<=K|t_n|<=M

    Thus,

    |t_n|<=M/K

    Thus,

    {t_n} must be a bounded sequence.
    Q.E.D.
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  4. #4
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    Quote Originally Posted by ThePerfectHacker View Post
    Theorem A convergent sequence is bounded.

    We have shown that,

    lim (s_n*t_n)=0

    Thus,

    |s_n*t_n|<= M

    By the previous theorem.

    BUT! {s_n} is also bounded by some non-zero konstant K because it is convergent.

    Thus,

    |s_n*t_n|<=K|t_n|<=M

    Thus,

    |t_n|<=M/K

    Thus,

    {t_n} must be a bounded sequence.
    Q.E.D.

    After I posted this yesterday I realized I made a mistake.
    I will not tell you were, you shall need to find it yourself.

    But what you said is false.
    Consider,
    s_n=0.
    And
    t_n=n

    Note,
    lim (s_nt_n)=0.
    But {t_n} is not bounded!

    I will state the following as an excercise (a little hard to show). But try it.

    Theorem Let lim (s_n) not = 0 and s_n not =0. Let t_n be any sequence. If {s_n*t_n} is a convergent sequence, then {t_n} must be bounded.
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