Show that If $\sum_{n=1}^\infty b_{n}$ is abosolutely convergent, then $|\sum_{n=1}^\infty b_{n}| \leq | \sum_{n=1}^\infty |b_{n}|$ this problem was given as a practice problem for first year calculus class. Here it is: show that if the series $\sum_{n=1}^\infty b_{n}$ is abosolutely convergent, then $|\sum_{n=1}^\infty b_{n}| \leq | \sum_{n=1}^\infty |b_{n}|$. Does this solutions work?
$\displaystyle Q_{k}=\sum_{n=1}^k b_{n}$ and $\displaystyle T_{k}=\sum_{n=1}^k |b_{n}|$ $\;$ and let $\displaystyle S=\sum_{n=1}^{\infty}b_{n}$ and $\displaystyle Q=\sum_{n=1}^{\infty}|b_{n}|$; 
so $S=\displaystyle\lim_{k\to\infty}S_{k}$ and $Q=\displaystyle\lim_{k\to\infty}Q_{k}$.
Then, $\;\;|S_{k}|=|b_{1}+ b_{2} + b_{3}\cdots+b_{k}|\le|b_{1}|+ |b_{2}| + |b_{3}|\cdots+|b_{k}|=Q_{k}$ and therefore 
$\;\;\;-T_{k}\le S_{k}\le T_{k}$ $\;\;$for $k\ge1$.
 A: You know that if we have two convergent sequences (in the wide sense of the word) $\;\{a_n\}\;,\;\;\{b_n\}\;$ s.t. $\;a_n\le b_n\;$ , then $\;\lim a_n\le\lim b_n\;$ .
Well, now apply this to the sequences of partial sums by means of the triangle inequality:
$$\forall\,N\in\Bbb N\;,\;\;\left|\sum_{n=1}^na_n\right|\le\sum_{n=1}^N|a_n|\implies \lim_{N\to\infty}\left|\sum_{n=1}^na_n\right|\le\lim_{N\to\infty}\sum_{n=1}^N|a_n|$$
and we're done .
You did almost the above, yet there seems to be a confusion with that misterious $\;S_k\;$ that appears there (apparently, it should be $\;S_k=T_k\;$), but it never minds: I think this makes things messier and the above seems to me simpler and shorter.
A: Alternatively you can show that $(s_n)$ [the $n$-th partial sum]  is a Cauchy sequence: 
Denote $S_n$ the $n$-th partial sum for the absolute values of $b_n$, i.e., $S_N =\sum_{n=0}^N |b_n|$. Choose $n_0$ such that $S_q-S_p<\varepsilon$,  for all $q>p\ge n_0$. Then 
$$|s_q-s_p|=\bigg|\sum_{n=p+1}^q b_n \bigg|\le\sum_{n=p+1}^q |b_n|=S_q-S_p< \varepsilon$$
Hence $(s_n)$ is Cauchy sequence and then converges, as desired.
