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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?

1 Answers 1

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  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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    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 $n\rightarrow \infty$.2012-10-14