TITLE: Constructing a Convergent/Divergent Series from a Positive Sequence
QUESTION [7 upvotes]: Suppose that $\{x_n\}$ is a positive sequence such that $x_n \to 0$. Construct a positive sequence $s_n$ such that $\displaystyle \sum_{n=1}^\infty s_n$ diverges while $\displaystyle \sum_{n=1}^\infty x_ns_n$ converges.
It is clear that if $\sum x_n$ converges, then taking $s_n=1$ for all $n$ will work. We are then reduced to the case where $\sum x_n$ diverges. My only thought is that as $\sum x_n$ does not converge, its tail does not form a Cauchy sequence, hence for some $\epsilon>0$, we have
$$
\sum_{n=m}^\infty x_n >\epsilon
$$
Any hint as to how I should proceed?

REPLY [3 votes]: Because $x_n\to 0,$ we can say i) the $x_n$'s are bounded above by some $M;$ and ii) There are $n_1 < n_2 < \cdots $ such that $x_{n_k} < 1/k^2$ for each $k.$
Let $E = \{n_1,n_2,\dots \}.$ For $n \in E,$ define $s_n = 1.$ For $n\notin E,$ define $s_n=1/n^2.$ Because $s_n = 1$ for infinitely many $n,$ $\sum s_n =\infty.$ We have
$$\sum_{n=1}^{\infty} x_ns_n = \sum_{n\in E} x_ns_n + \sum_{n\notin E} x_ns_n.$$
The first sum on the right is 
$$\sum_{k=1}^\infty x_{n_k}s_{n_k}< \sum_{k=1}^\infty \frac{1}{k^2}\cdot 1 < \infty.$$
The second sum on the right is no more than
$$\sum_{n\notin E} M\cdot \frac{1}{n^2} < \infty.$$
Thus $\sum_{n=1}^{\infty} x_ns_n$ converges as desired.