Convergence of a product series If $\sum a_n$ and if $\{b_n\}$ is monotonic and bounded, I want to prove that $\sum a_nb_n$ converges. There is a theorem in the book of Rudin which gives sufficient condition for a product series to converge. But, these conditions are not satisfied in this case. How to proceed then ? 
 A: Assuming $\sum a_n$ is convergent and $(b_n)$ is monotonic and bounded, we can always add a constant to $b_n$ and multiply by $-1$ if necessary to make $(b_n)$ increasing and nonnegative. Now define $$s_n=\sum_{k=1}^n a_n,\quad c_n=b_n-b_{n-1}$$ (and say $s_0=b_0=0$ for convenience), then use partial summation:
$$\begin{aligned} s_nb_n
&=\sum_{k=1}^n (s_kb_k-s_{k-1}b_k+s_{k-1}b_k-s_{k-1}b_{k-1})\\
&=\sum_{k=1}^n a_kb_k-\sum_{k=1}^ns_{k-1}c_k
\end{aligned}
$$
(I always have to write that out to avoid one-off errors).
Here the left hand side converges as $n\to\infty$, and so does the second sum on the right, for $\sum c_k$ is absolutely convergent and $(s_{k-1})$ is bounded. Thus the first sum on the right also converges.
A: Besides Herald good answer here is a way using the defintion of a limit.
Since $a_n$ and $b_n$ are bounded then we have that $b_n \le K$ for some $K\in \mathbb{R}$ and by Cauchy, given an $\epsilon >0$ there exists an $N$ such that for all $n>m> N$ we have that $$|S_n - S_m| = ||a_n|-|a_{n-1}| + ... + |a_{m+1}|| < \delta$$ where delta we will determine later since it is not obvious now what it should be. $S_n$ is the $n^{th}$ partial sum of $\sum_{n=1}^\infty |a_n|.$ Thus we have for the same $N$ and for all $m,n >N$ $$|a_nb_n -a_mb_m| = |a_nb_n+a_{n-1}b_{n-1} + ... + a_{m+1}b_{m=1}| \le |a_nb_n| + |a_{n-1}b_{n-1}| + ... + |a_{m+1}b_{m+1}| \le K(|a_n| +|a_{n-1} + ...+ |a_{m+1}|)< K\delta<\epsilon$$ thus $\delta$ should equal $\delta = \frac{\epsilon}{K}$. 
