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$\{a_n\}$ converges. Can $\{|a_n|\}$ diverge?

My try:

$\{a_n\}$ converges $\Longleftrightarrow$ $\exists l\in R,\lim_{n\to\infty} a_n = l$

$\lim_{n\to\infty} a_n = l \Longrightarrow \lim_{n\to\infty} |a_n| = |\lim_{n\to\infty} a_n| = |l|$

And now we prove the previous line:

$\lim_{n\to\infty} a_n = l \Longleftrightarrow \forall\epsilon > 0 , \exists n_o\in N , \forall n>n_0 : |a_n-l|<\epsilon$

We know by the triangle inequality that:

$\left||a_n|-|l|\right|\leq |a_n-l|<\epsilon$

Then:

$\forall\epsilon > 0 , \exists n_o\in N , \forall n>n_0 : ||a_n|-|l||<\epsilon\Longleftrightarrow \lim_{n\to\infty} |a_n| = |l| $

This means that the answer to the question is NO

I am aware that this is similar to this other post , but it didnt solve my problem.

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    $\begingroup$ That's right. In other words $x\mapsto|x|$ is continuous on $\Bbb R$. $\endgroup$ – Lord Shark the Unknown Oct 22 '18 at 17:22
  • $\begingroup$ To take the limit inside a function it only needs to be continuous? Doesnt it have to be continuous AND derivable? $\endgroup$ – user605734 MBS Oct 22 '18 at 17:28
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    $\begingroup$ No, the function only needs to be continuous. $\endgroup$ – Clayton Oct 22 '18 at 17:48
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As an alternative, suppose by contradiction that $|a_n|$ diverges therefore since

  • $|a_n|=a_n$ for $a_n\ge 0$

  • $|a_n|=-a_n$ for $a_n< 0$

we have that $a_n$ doesn’t converge.

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