# If $\sum a_n$ converges, and if {$b_n$} is monotonic and bounded, prove $\sum a_n b_n$ converge.

I proved this way:

Since {$b_n$} is bounded, let $|b_n| \le \alpha$ and $\alpha$ is an upper bound.
Since $\sum a_n$ converges, there exists N such that $|\sum_{k=m}^{n} a_k| \le \frac{\epsilon}{\alpha}$ for every N $\le$ n,m.
Then $|\sum_{k=m}^{n} a_k \alpha| \le \epsilon$ for every N $\le$ n,m.
Now, $|\sum_{k=m}^{n} a_k b_k| \le |\sum_{k=m}^{n} a_k \alpha| \le \epsilon$, thus $\sum a_n b_n$ converges.

But the professor said the last inequality is wrong. What is the problem? How to fix it?

• The same problem was discussed in this question. (But it is not a duplicate, since here the OP asks whether there is a problem with his proof.) – Martin Sleziak Nov 20 '12 at 8:27

$\def\abs#1{\left|#1\right|}$The problem is, that you in general can only say $$\abs{\sum_{k=m}^n a_k b_k} \le \sum_{k=m}^n \abs{a_k}\abs{b_k} \le \alpha \cdot \sum_{k=m}^n \abs{a_k}$$ but this doesn't help you here as $\sum_k a_k$ need not to be absolutely convergent. You cannot pull out an upper bound if you do not know about the signs of the summands being the same. Consider the following example: $a_1 = 2$, $a_2 = -2$, $b_1 = 3$, $b_2 = 1$. Then $\alpha = 3$ is an upper bound of the $b_k$, and we have \begin{align*} \abs{a_1b_1 + a_2b_2} &= \abs{2 \cdot 3 - 2 \cdot 1}\\ &= \abs 4\\ &\not\le \alpha\abs{a_1 + a_2}\\ &= 3 \cdot \abs{2 - 2}\\ &= 0. \end{align*} For a solution to your given problem, see this question here.