Does the comparison test imply absolute convergence?

Question

As my textbook states it:

Let $\Sigma a_n$ be a series where $a_n \ge 0$ for all $n $.

(i) If $\Sigma a_n $ converges and $|b_n| \le a_n$ for all $n $, then $\Sigma b_n$ converges.

(ii) If $\Sigma a_n = +\infty $ and $b_n \ge a_n $ for all $n $, then $ \Sigma b_n = +\infty$

My question is, does this imply absolute convergence for $\Sigma b_n$, or just convergence? It seems to me that this implies absolute convergence, as

$$|\Sigma b_n| \le \Sigma |b_n| = |\Sigma |b_n|| \le \Sigma a_n $$ So both $\Sigma b_n $ and $\Sigma |b_n|$ satisfy the Cauchy criterion.

Answer 1

Yes, this implies that $\sum|b_n|$ converges (and hence that $\sum b_n$ converges absolutely). Let $c_n=|b_n|$. We know that $|c_n|=\big||b_n|\big|= |b_n|\le a_n$ for all $n$; then by (i), $\sum c_n$ converges.