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Math Help - Convergent Sequence Inequality

  1. #1
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    Convergent Sequence Inequality

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

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

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

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


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