Convergence of a sequence in absolut value.

Question

I need to prove this:

If $a_{n}$ converges to $A$, then $|a_{n}|$ converges to $|A|$.

And I have this:

$a_{n} \rightarrow A$ then, given $\epsilon>0$ there exists $N \in J$ such that $$|a_{n}-A|<\epsilon$$ for $n \ge N$. Now, we consider the sequence: $$\{|a_{n}| \}$$ And we claim that: $$\{|a_{n}| \}\rightarrow |A|$$

Then using the reverse triangle inequality we have that: $$\epsilon>|a_{n}-A|>||a_{n}|-|A||\\ \Rightarrow\epsilon>||a_{n}|-|A||$$

then we have that $\{|a_{n}| \}\rightarrow |A|$

Am I right?, and Is the converse true?, Why?. Thank you.

Answer 1

You're right: just correct some typo! The sequence $(|a_n|)$ converges to $|A|$ if $(a_n)$ is convergent to $A$.

The converse isn't true: consider the sequence $((-1)^n)$.

Answer 2

You got it $99.99\%$ right: just note that $|a_n-A|\geq ||a_n|-|A||$ (note the weak inequality).

The converse is false: the alternating sequence $1,-1,1,-1,\ldots$ diverges but converges absolutely to $1$.