Too simple proof for convergence of $\sum_n a_n b_n$? My question relates to Chapter 3, Exercise 8 in "Baby Rudin". It states:
If $\sum_n a_n$ converges, and if $\{b_n\}$ is monotonic and bounded, prove that $\sum_n a_n b_n$ converges.
My attempt would have been:
Since $\{b_n\}$ is monotonic and bounded, $\{b_n\}$ converges and it exists $\inf \{b_n\}$ as well as $\sup \{b_n\}$. But then we have
$$ \left| \sum_n a_n \inf \{b_n\} \right| \leq \left| \sum_n a_n b_n \right| \leq  \left| \sum_n a_n \sup \{b_n\} \right| \leq \max \left( \left| \sup \{b_n\} \right|, \left| \inf \{b_n \} \right| \right)\varepsilon \leq \tilde{\varepsilon} $$
since $\sum_n a_n$ converges and $\max \left( \left| \sup \{b_n\} \right|, \left| \inf \{b_n \} \right| \right)$ is a finite number. This would imply, by the comparison test, that $\sum_n a_n b_n$ converges as well. $\quad \Box$
But as there was a way longer, more rigorous proof chosen in this solution manual, I'm a bit suspicious that my proof is not complete. Am I missing something?

EDIT: Thanks everyone! @hermes:
The last part of your proof gave me the following idea. As $\lim_{n \to \infty} b_n = c$ and $\{b_n\}$ is monotonic, couldn't we just set $b_n = c - c_n$ with a monotonically decreasing sequence $\{c_n\}$ which has $\lim_{n \to \infty} c_n = 0$. Then we have
$$\sum_n a_n b_n = \underbrace{c \sum_n a_n}_{\text{converges by assumption}} - \underbrace{\sum_n a_n c_n}_{\text{converges by Theorem 3.42}} \leq \varepsilon_1 - \varepsilon_2 = \varepsilon $$
since

Theorem 3.42 Suppose

*

*the partial sums of $\sum_n a_n$ form a bounded space $\quad \checkmark$

*$c_0 \geq c_1 \geq c_2 \geq \dots \quad \checkmark$

*$\lim_{n \to \infty} c_n = 0 \quad \checkmark$
Then $\sum_n a_n c_n$ converges.

Thus $\sum_n a_n b_n$ converges as well. $\quad \Box$
Now that should hold, I think. So I wouldn't need to go through all the estimates.
 A: As Daniel Fischer points out, if the $a_n$ don't all have the same sign, your inequalities don't necessarily hold. You should use partial summation (used in proving Abel theorem) to get an estimate of Cauchy sum.
Let $A_n=\sum_{k=m}^n a_k$. So by Cauchy Criterion, $|A_n|<\epsilon$ for $n,m>N$.
We have
\begin{align}
\sum_{k=m}^n a_kb_k&=\sum_{k=m}^n (A_k-A_{k-1})b_k
\\
&=\sum_{k=m}^n A_kb_k -\sum_{k=m}^n A_{k-1}b_k
\\
&=\sum_{k=m}^{n-1} A_k(b_k-b_{k+1})+A_nb_n\tag{$A_{m−1}=0$}
\end{align}
First suppose $\lim_{n\to\infty}b_n=0$ and $\:b_n \downarrow$. Then $\:b_n\geqslant0$, and $b_k-b_{k+1}\geqslant0\:$ for all $k>0$.
Since $-\epsilon<A_k<\epsilon$ for all $k>m$
$$
|A_k(b_k-b_{k+1})|<\epsilon(b_k-b_{k+1})
$$
So for all $n,m>N-1$, there is
\begin{align}
\left|\sum_{k=m}^n a_kb_k\right|&\leqslant\sum_{k=m}^{n-1} |A_k(b_k-b_{k+1})|+|A_nb_n|
\\
&\leqslant\sum_{k=m}^{n-1} \epsilon\:(b_k-b_{k+1})+\epsilon \:b_n
\\
&=\epsilon \:(b_m-b_n+b_n)
\\
&=\epsilon \:b_m
\\
&\leqslant M\epsilon
\end{align}
So by Cauchy Criterion, $\sum_{k=1}^{\infty} a_kb_k$ converges. 
Finally if $\lim_{n\to\infty}b_n=c\ne0$, we replace $b_n$ with $b_n-c$. Then
$$
\sum_{k=1}^{\infty} a_kb_k=\sum_{k=1}^{\infty} a_k(b_k-c)+c\sum_{k=1}^{\infty} a_k 
$$
converges for $\sum_{k=1}^{\infty} a_k(b_k-c)$ converges.
