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Math Help - Limit Proof

  1. #1
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    Limit Proof

    Show that a_n > 0 ,\ \forall n \in \mathbb{N} that \lim_{n \rightarrow \infty} a_n = 0 if and only if \lim_{n \rightarrow \infty} \frac{1}{a_n} = \infty.

    Attempt:
    \lim_{n \rightarrow \infty} a_n = 0 \Rightarrow \lim_{n \rightarrow \infty} \frac{1}{a_n} = \infty.

    so then we have |a_n - 0| < \epsilon, \ \forall \epsilon > 0 which would simply imply |a_n| < \epsilon, \ \forall \epsilon > 0

    If I use the definition:

    Let \alpha be in \mathbb{R}. The open ray (\alpha, \ \infty) =\{x \ \in \ \mathbb{R}: \ x>\alpha\}

    so I would get something along the lines of:
    \frac{1}{|a_n|} > \frac{1}{\epsilon} then I would have

    \alpha < \frac{1}{|a_n|} we can see that \frac{1}{|a_n|} tends to \infty

    \lim_{n \rightarrow \infty} \frac{1}{a_n} = \infty \Rightarrow \lim_{n \rightarrow \infty} a_n = 0.

    \left| \frac{1}{a_n} -\alpha \right| < \epsilon \ \ \ \ \ \ \forall \ \epsilon >0  this would imply:

    -\epsilon < \frac{1}{a_n} -\alpha < \epsilon \ \ \ \ \ \ \forall \ \epsilon >0

    (\alpha-\epsilon) \cdot a_n < 1 < (\alpha+\epsilon) \cdot a_n \ \ \ \ \ \ \forall \ \epsilon >0

    at which point I get stuck
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  2. #2
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    Here is a very simple outline.
    If \left( {a_n } \right) \to 0 then \left( {\forall K \in \mathbb{Z}^ +  } \right)\left( {\exists n_K } \right)\left[ {0 < a_{n_K }  < \frac{1}{K}} \right]\; \Rightarrow \;K < \frac{1}{{a_{n_K } }}.

    If \left( {\frac{1}{{a_n }}} \right) \to \infty then \left( {\forall K \in \mathbb{Z}^ +  } \right)\left( {\exists n_K } \right)\left[ {K < \frac{1}{{a_{n_K } }}} \right]\; \Rightarrow \;a_{n_K }  < \frac{1}{K}.
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