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Let $\{a_n\}$ be a sequnce. Then $a_n \to -\infty$ if $\forall K < 0 \;\exists N \;\forall n \ge N:a_n < K$

Show that:

  1. If $a_n → -\infty$, $a_n \ne 0$, then $1/a_n→0$ ; and

  2. If $a_n < 0$, $a_n → 0$, then $1/a_n→−∞$

From the highschool I know that this is true. But do not know how to prove it. Can you help me?

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1 Answer 1

  1. Fix $\epsilon > 0$

    Then there exists, $N\in \mathbb{N}$ such that $n\geq N \Rightarrow a_n < -1/{\epsilon}$

    Thus, $n\geq N \Rightarrow 0> 1/a_n > -\epsilon$

    This shows that $a_n \rightarrow 0$

  2. Fix $\epsilon < 0$

    Then there exists, $N\in \mathbb{N}$ such that $n\geq N \Rightarrow |a_n| <-1/{\epsilon}$

    Since, $a_n < 0$, $n\geq N \Rightarrow \epsilon > 1/{a_n}$

    Consequently, $1/{a_n} \rightarrow -\infty$

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what I do not get is how can you conclude that an→0 if 0>1/an>−ϵ. I know that ϵ is a small number but that does not make sense for me to write directly an→0. –  alev Oct 14 '12 at 21:55
    
@alev Check that I 'fixed' $\epsilon > 0$. Since statement holds for arbitrary $\epsilon >0$, the statement must be true for every $\epsilon >0$. I didn't mean $\epsilon$ to be a small number, but arbitrary real number. If you still don't understand, let me know :) –  Rubertos Oct 14 '12 at 22:17
    
I think I am not confident with the part that "0>1/an>−ϵ this shows that an→0" You are not skipping any steps in the proof right? Sorry for the inconvenience :/ –  alev Oct 14 '12 at 22:27
    
Thus 'For every $\epsilon > 0$, there exists $N\in \mathbb{N}$ such that $n \geq N \Rightarrow 0>1/{a_n}>-\epsilon$' is true. By the definition of limit, $a_n \rightarrow 0$ as $n\rightarrow \infty$. –  Rubertos Oct 14 '12 at 22:27
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