Why doesn't this work for Rudin Exercise 3.8 The problem is 3.8 exercise in baby Rudin:

If $ \sum{a_n} $ converges and $\{b_n\}$ is bounded and monotonic, prove that $\sum{a_nb_n}$ converges. 

Why can't I just do this?:
Let $M$ be an upper bound of $\{b_n\}$. Choose $N$ such that for all $n,m\ge N$,
$$ \sum_{k=n}^m{a_k} \le {\epsilon \over  M}$$ 
Then, $$\sum_{k=n}^m{a_kb_k} \le M\sum_{k=n}^m{a_k} \le \epsilon $$
If someone has a quickish clear proof for this, I'd love to see it also. Thanks in advance. 
 A: While the proper way to prove the convergence of $\sum a_{n}b_{n}$ is the one provided by robjohn's answer, I would like to point out flaw in your argument. You start off correctly but you need to use the absolute values. Thus if $|b_{n}| \leq M$ for all $n$ and $\epsilon > 0$ we can choose a positive integer $n_{1}$ such that for all integers $m > n \geq n_{1}$ we have $$\left|\sum_{k = n}^{m}a_{k}\right| < \frac{\epsilon}{M}\tag{1}$$ Note that this is not the same thing as saying $$\sum_{k = n}^{m}|a_{k}| < \frac{\epsilon}{M}\tag{2}$$ Equation $(2)$ implies equation $(1)$ but $(1)$ does not imply $(2)$ because of the standard triangle inequality $$\left|\sum_{k = n}^{m}a_{k}\right| \leq \sum_{k = n}^{m}|a_{k}|\tag{3}$$ Now we can see that $$\left|\sum_{k = n}^{m}a_{k}b_{k}\right| \leq \sum_{k = n}^{m}|a_{k}b_{k}| \leq M\sum_{k = n}^{m}|a_{k}|$$ and your argument would work if equation $(2)$ holds. But unfortunately the equation $(1)$ holds instead of the much needed equation $(2)$ and therefore your argument is flawed and does not lead to a correct proof.
A: Need to Use Monotonicity
Since you did not use the monotonicity of $b_n$, you would need to know that the sum converges absolutely to make your argument work. Take for example
$$
a_n=\frac{(-1)^{n-1}}n
$$
and
$$
b_n=1+(-1)^{n-1}
$$
The sum of $a_n$ is a well known convergent series:
$$
\sum_{n=1}^\infty a_n=\sum_{n=1}^\infty\frac{(-1)^{n-1}}n=\log(2)
$$
However, since $b_n=0$ for even $n$ and $b_n=2$ for odd $n$, we have that
$$
\sum_{n=1}^\infty a_nb_n=\sum_{n=1}^\infty\frac2{2n-1}
$$
which diverges by comparison to the Harmonic series.

Using Monotonicity
Since $b_n$ is bounded and monotonic, let $b_\infty=\lim\limits_{n\to\infty}b_n$. Furthermore, since $\sum\limits_{n=1}^\infty a_n$ converges, it is bounded independent of $n$.
Thus, we have that
$$
\begin{align}
\sum_{n=1}^\infty a_nb_n
&=\underbrace{\sum_{n=1}^\infty a_n(b_n-b_\infty)}_{\text{converges by Dirichlet's Test}}+\underbrace{b_\infty\sum_{n=1}^\infty a_n}_{\substack{\text{constant multiple of}\\\text{a convergent series}}}
\end{align}
$$
converges.
