Prove that if $\lim \{a_n\}\to a$ then $\lim \{|a_n|\}\to|a|$. Is the converse true?

I don't know where to start. Should I break this into 2 cases?

  • $\begingroup$ you can go back to the definition of limit, with $\epsilon$. $\endgroup$
    – Denis
    Sep 11, 2013 at 20:38
  • 4
    $\begingroup$ What do you think about $a_n=(-1)^n$ $\endgroup$
    – Bertrand R
    Sep 11, 2013 at 20:39
  • $\begingroup$ What do you mean? The limits as $n \to \infty$? $\endgroup$ Sep 11, 2013 at 20:39
  • $\begingroup$ In addition to dkuper's comment, you are probably missing a property of $|.|$ : $||a|-|b||\leq |a-b|$ $\endgroup$
    – Bertrand R
    Sep 11, 2013 at 20:43

1 Answer 1

  • By definition, if $$\lim_{n \rightarrow \infty} a_n=a$$ then for all $\varepsilon>0$ there exists a $N \in \mathbb{N}$ such that $n \geq N$ implies $|a_n-a| \leq \varepsilon$.

  • We may also verify case-by-case that $$\big||x|-|y|\big| \leq |x-y|$$ for all $x,y \in \mathbb{R}$. In particular,$$\big||a_n|-|a|\big| \leq |a_n-a|.$$

Combining these yields the proof.


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