Show that if $\Sigma$ $a_n$ converges absolutely, then $\Sigma$ $a^2_n$ also converges. Is this still true if $\Sigma$ $a_n$ converges conditionally?

Question

Do I use a convergence test to solve this? The definition of converges absolutely is $\Sigma_{n = 1}^{\infty}$|$a_n$|

Answer 1

Hint :

Since $a_n\to 0$, there is an $N$ s.t. $$|a_n|<1$$ if $n>N$, and thus $$|a_n|^2<|a_n|$$ when $n>N$.

Answer 2

No, $\sum\frac {(-1)^n}{\sqrt n}$ converges conditionally, but it is well-know that $\sum \frac 1n$ does not.