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Math Help - Another divergence proof: Prove 1/a_n converges to ∞ (i.e. 1/a_n diverges).

  1. #1
    Member ilikedmath's Avatar
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    Exclamation Another divergence proof: Prove 1/a_n converges to ∞ (i.e. 1/a_n diverges).

    If a_n > 0 \forall n \in \mathbb{N} and \lim_{n\rightarrow\infty} a_n = 0, prove that \dfrac{1}{a_n} \rightarrow \infty.

    I haven't been able to write up a proof for this one yet. I have a few ideas down:

    Since { a_n} \rightarrow 0, by definition of convergence, I know | a_n| < \varepsilon, or
    - \varepsilon< a_n< \varepsilon.

    Starting ideas for proof:
    Assume \lim_{n\rightarrow\infty} a_n = 0 and a_n > 0 \forall n \in \mathbb{N}.
    { a_n} \rightarrow 0 \Rightarrow| a_n - 0| < \varepsilon \Rightarrow | a_n| < \varepsilon \Rightarrow | \dfrac{1}{a_n}| > \dfrac{1}{\varepsilon}, since \dfrac{1}{\varepsilon} > 0 .....

    I was working with my classmates, and we were guessing that we would reach a contradiction somehow using the definition of convergence, and then by ______ (we don't know what exatly) we will conclude that the sequence has to diverge by contradiction to our assumption.

    Any help, suggestions, tips, corrections, guidance, etc. is greatly appreciated. Thank you very much for looking and for your time!
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  2. #2
    Moo
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    Hello,

    Just use the definitions

    a_n \to 0 ~:

    \forall \varepsilon >0,~ \exists N \in \mathbb{N},~ \forall n \geq N,~ |a_n|=a_n< \varepsilon

    The inequality can be rewritten :
    \frac{1}{a_n} > \frac 1 \varepsilon

    So finally, we have :
    \forall A>0,~ \exists N \in \mathbb{N},~ \forall n \geq N,~ \frac{1}{a_n} > A
    where in fact A=\frac 1 \varepsilon, but it doesn't matter since we take random \varepsilon


    And this is the exact definition of \frac{1}{a_n} \to +\infty


    If you can't visualize it, take b_n=\frac{1}{a_n} and rewrite the definition of "limit goes to + infinity" for b_n.
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  3. #3
    Member ilikedmath's Avatar
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    Quote Originally Posted by Moo View Post
    Hello,

    Just use the definitions
    Awesome. Thanks!
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