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. 

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.
