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Thread: Convergent Sequence Inequality

  1. #1
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    Convergent Sequence Inequality

    If $\displaystyle a_n\,$ is a convergent sequence and $\displaystyle \lim a_n >a\,$. Show that $\displaystyle a_n > a\,$ for some $\displaystyle n>N\,$.

    My attempt was like this.
    I used the limit limitation theorem*.

    Assume $\displaystyle a_n\leq a$ for sufficiently large $\displaystyle n\,$
    Then, $\displaystyle \lim a_n\leq a$
    A contradiction.

    Thus, the statement $\displaystyle a_n\leq a$ for sufficiently large $\displaystyle n\,$ is false.
    My difficutly is that we cannot conclude that $\displaystyle a_n>a$.
    Because that cannot be the negation of the statement.
    It may mean that, there is no such $\displaystyle N\,$ such
    $\displaystyle a_n\leq a$ for $\displaystyle n>N$.
    Meaning we can have,
    $\displaystyle a_n>a$ for $\displaystyle n=N+1$.
    $\displaystyle a_n\leq a$ for $\displaystyle n=N+2$.
    $\displaystyle a_n > a$ for $\displaystyle n=N+3$.
    And thus on....


    *)If $\displaystyle a_n\,$ is a convergent sequence and $\displaystyle a_n\geq a\,$ for sufficiently large $\displaystyle n\,$ then $\displaystyle \lim a_n\geq a\,$.
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  2. #2
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    For notation sake say $\displaystyle \left( {a_n } \right) \to L > a$.
    Then we have:
    $\displaystyle \begin{array}{rcl}
    \frac{{L - a}}{2} > 0\quad & \Rightarrow & \quad \left( {\exists K \in Z^ + } \right)\left[ {\left| {a_K - L} \right| < \frac{{L - a}}{2}} \right] \\
    \quad & \Rightarrow & \quad - \frac{{L - a}}{2} < a_K - L \\
    & \Rightarrow & \quad L - \frac{{L - a}}{2} < a_K \\
    & \Rightarrow & \quad a < \frac{{L + a}}{2} < a_K \\
    \end{array}$
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  3. #3
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    Quote Originally Posted by Plato View Post
    For notation sake say $\displaystyle \left( {a_n } \right) \to L > a$.
    Then we have:
    $\displaystyle \begin{array}{rcl}
    \frac{{L - a}}{2} > 0\quad & \Rightarrow & \quad \left( {\exists K \in Z^ + } \right)\left[ {\left| {a_K - L} \right| < \frac{{L - a}}{2}} \right] \\
    \quad & \Rightarrow & \quad - \frac{{L - a}}{2} < a_K - L \\
    & \Rightarrow & \quad L - \frac{{L - a}}{2} < a_K \\
    & \Rightarrow & \quad a < \frac{{L + a}}{2} < a_K \\
    \end{array}$
    Thank you, excellent job.
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