# Show that $\sum a_n b_n$ is convergent

If $\sum a_n$ is convergent and $b_n$ is a monotone bounded sequence then $\sum a_nb_n$ is convergent.

What I did is

If $b_n$ is a monotone bounded sequence then $b_n$ is convergent and suppose that $b_n\rightarrow B$ when $b\rightarrow\infty$.

The algebraic limit theorems says that $\sum a_n B$ is convergent. Since $0\leq a_n b_n\leq a_n B$ then by the comparison test $\sum a_n b_n$ is also convergent.

It's a wrong use of comparison test?

• Why is $0 \leq a_n b_n$? – Zain Patel Jun 9 '17 at 18:20
• I think it's necessary that both $\Sigma a_n$ and $\Sigma b_n$ be absolutely convergent for you to infer that $\Sigma a_n b_n$ is convergent. – Ryan A Jun 9 '17 at 18:26
• It doesn't even follow that $a_nb_n \leq a_nB$, consider $b_n = \frac{-1}{n^2}$, or really any other convergent series with negative partial sums. – Duncan Ramage Jun 9 '17 at 18:28
• @Ryan not really -- en.wikipedia.org/wiki/Abel%27s_test – Clement C. Jun 9 '17 at 18:28
• @HaraldHanche-Olsen that's indeed the standard way to do this. – Zain Patel Jun 9 '17 at 18:28

It is enough to apply Dirichlet's test. If $\sum_{n\geq 1}a_n$ is convergent to $A$ then its partial sums are bounded, and if $\{b_n\}_{n\geq 1}$ is monotonic and bounded it is convergent to some $B$ and $B-b_n$ is decreasing to zero. It follows that

$$\sum_{n\geq 1}a_n b_n = A B - \underbrace{\sum_{n\geq 1} a_n (B-b_n)}_{\text{convergent by Dirichlet}}$$ is convergent as well.