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

$$\{a_n\}$$ 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.

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

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.