If $a_n\rightarrow a$ then $|a_n|\rightarrow |a|$ why does it not work for series?

Question

Proof:

Let $\epsilon>0$ . If $a_n\rightarrow a$ there exists a $N$ such that for every $n>N$ we have $|a_n-a|<\epsilon$

For the same $n$ we have

$||a_n|-|a||\leq|a_n-a|<\epsilon$

Therefore $|a_n|\rightarrow |a|$

But if $s_n=\sum_{k=1}^n(-1)^kk^{-1}$ then $s_n\rightarrow \ln(2)$

But $|s_n|=\sum_{k=1}^nk^{-1}\not\rightarrow \ln(2)$

The textbook says that it works for

$\sum_n \overline {a_n}=\overline{\sum_n a_n}$

But why does not work this Argumentation for $|\cdot|$ ?

Answer 1

Your assertion that $$|s_n|=\sum_{k=1}^nk^{-1}$$ is plain false. It it not true in general that $$\left|\sum_{k=1}^n a_k\right| = \sum_{k=1}^n \left|a_k\right|$$