Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
  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, $1/{a_n} \rightarrow -\infty$

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.