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?
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you can go back to the definition of limit, with $\epsilon$. – Denis Sep 11 '13 at 20:38
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4What do you think about $a_n=(-1)^n$ – Bertrand R Sep 11 '13 at 20:39
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What do you mean? The limits as $n \to \infty$? – Étienne Bézout Sep 11 '13 at 20:39
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In addition to dkuper's comment, you are probably missing a property of $|.|$ : $||a|-|b||\leq |a-b|$ – Bertrand R Sep 11 '13 at 20:43
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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.
Rebecca J. Stones
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2Maybe it is worth mentioning that this inequality is called [reverse triangle inequality](http://en.wikipedia.org/wiki/Reverse_triangle_inequality) and some proofs can be found also [here on MSE](http://math.stackexchange.com/questions/127372/reverse-triangle-inequality-proof). – Martin Sleziak Sep 02 '14 at 11:04