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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?

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  • $\begingroup$ This might be total BS, but: suppose $\{x_n\}$ were a $p$-series, that is $x_n = n^p$. If $x_n \rightarrow 0$ then $p < 0$. Then we can choose choose $s_n$ such that it "makes up the difference", i.e. $s_n = n^{-1 - p - \epsilon}$, so $s_n$ doesn't converge (because $p - 1 - \epsilon > -1$) but $x_n s_n = n^{p - 1 - p - \epsilon} = n^{-1 - \epsilon}$ and it converges. $\endgroup$ – Eli Rose -- REINSTATE MONICA Jan 6 '16 at 6:26
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    $\begingroup$ Hint: for each $k\ge0$, let $N_k$ be the set of indices $n$ such that $x_n \in (2^{-(k+1)}, 2^{-k}]$; define $s_n$ to be the same for all elements $n\in N_k$, with its value depending only on $\#N_k$. $\endgroup$ – Greg Martin Jan 6 '16 at 7:00
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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.

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