Little Rudin series convergence exercise Problem: If $\sum a_n$ converges, and $\{b_n\}$ is monotonic and bounded, prove that $\sum a_n b_n$ converges.
Source: Rudin, Principles of Mathematical Analysis, Chapter 3, Exercise 8.
 A: The sequence $\{b_n\}$ is monotonic and bounded, so it converges to some number $C$. Assume, without loss of generality, that the sequence $\{b_n\}$ is increasing, and write $b_n=C-d_n$, where $d_n\rightarrow 0$. We have 
$$\sum a_nb_n = C\sum a_n -\sum a_nd_n.$$
The first series on the right is convergent by hypothesis, and the second is convergent because of the following theorem:
Theorem: If the partial sums of $\sum t_n$ form a bounded sequence and $s_n$ is a decreasing sequence that tends to 0, then $\sum t_ns_n$ converges.
Here we take $t_n=a_n$ and $s_n=d_n$. 
A: I recently attempted this problem and came up with a solution which I don't think is as good as other answers here, but i want to know if it's correct or not. So here is my solution :
According to the hypothesis $\sum a_n$ converges and $\{b_n\}$ is bounded & monotonic. So there exists $T,M \gt 0$ such that $|A_n| = \left|\displaystyle\sum_{k=1}^{n} a_k\right| \le M$ and $|b_n| \le T$.
For every $\epsilon \gt 0$, there exists an $N_1 \gt 0$ such that $N_1 \le p \le q$ implies $\left|\displaystyle\sum_{n=p}^{q} a_n \right| \le \dfrac{\epsilon}{2T} \tag 1$
and again for every $\epsilon \gt 0$, there exists an $N_2 \gt 0$ such that $N_2 \le p \le q$ implies $|b_p - b_q| \le \dfrac{\epsilon}{4M} \tag2.$
$N = \text{max}(N_1, N_2)$
$\begin{align}
 \left|\displaystyle\sum_{n=p}^{q} c_n\right| = \left| \displaystyle\sum_{n=p}^{q} a_n 
 b_n \right| 
 &= \left| \displaystyle\sum_{n=p}^{q-1}A_n(b_n-b_{n+1}) + A_q b_q - A_{p-1} 
 b_p\right| \\
 &= \left| \displaystyle\sum_{n=p}^{q-1}A_n(b_n-b_{n+1}) + A_q(b_q - b_p) - (A_{p-1} - 
 A_q)b_p \right| \\
 &\le M\left| \displaystyle\sum_{n=p}^{q-1}(b_n-b_{n+1}) \right| + M\left| b_q - b_p 
 \right| + T\left| \displaystyle\sum_{n=p}^{q} a_n \right| \\
 &= 2M\left| b_q - b_p \right| + T\left| \displaystyle\sum_{n=p}^{q} a_n \right| \\
 &\le 2M\left( \dfrac{\epsilon}{4M} \right) + T\left( \dfrac{\epsilon}{2T} \right) = 
 \dfrac{\epsilon}{2} + 
 \dfrac{\epsilon}{2} = \epsilon 
\end{align}$
So basically this mean that for every $\epsilon \gt 0$ there exists an $N \gt 0$ such that $N \le p \le q$ implies $\left| \displaystyle\sum_{n=p}^{q} a_n 
 b_n \right| \le \epsilon.$
