Let $\{a_n\}_{n=1}^{\infty}$ be a sequence with non negative terms so that for every sequence $\{x_n\}_{n=1}^{\infty}$ with $x_n \geq 0$ and $\lim_nx_n=0$, the series $\sum_{n=1}^{\infty}x_na_n$ converges. Show that $ \sum_{n=1}^{\infty}a_n $ also converges.
We can proceed through contradiction by assuming $ \sum_{n=1}^{\infty}a_n = \infty$ and then constructing a sequence $\{x_n\}_{n=1}^{\infty}$ such that $ \sum_{n=1}^{\infty}x_na_n $ diverges.
If $\sum a_n=\infty$, there is an infinite sequence $\{n_k\}$ such that $\sum\limits_{n=n_{k-1}+1}^{n_k} a_n > 1$ for all $k$. Define $x_n=\frac{1}{k}$, where $n_{k-1} < n \leq n_k$.
I'll follow the poster's suggestion of assuming that the sum $\sum_n a_n=\infty$, and then construct a sequence $x_n$ so that $\sum_n x_na_n=\infty$.
We can assume without loss of generality that $a_n>0$ since we can otherwise just ignore those $n$ for which $a_n=0$.
Let $A_n=a_1+\cdots+a_n$ with $A_0=0$. By assumption, $A_n\rightarrow\infty$, and therefore so does $B_n=\sqrt{A_n}$. Then, the differences $b_n=B_n-B_{n-1}$ will also be positive with $\sum_n b_n=\infty$.
We now let $x_n=b_n/a_n$. This satisfies $x_n\approx 1/2\sqrt{A_n}$: i.e., the derivative of the square root transformation used to make $B_n$ from $A_n$. More precisely, $$ b_n=\sqrt{A_n}-\sqrt{A_{n-1}} =\frac{A_n-A_{n-1}}{\sqrt{A_n}+\sqrt{A_{n-1}}} =\frac{a_n}{\sqrt{A_n}+\sqrt{A_{n-1}}} $$ so that $$ x_n=\frac{b_n}{a_n}=\frac{1}{\sqrt{A_n}+\sqrt{A_{n-1}}}. $$
Since $A_n$ is strictly increasing towards infinity, $x_n$ will thus decrease towards zero and satisfy the requirements. However, $\sum_n x_na_n=\sum_n b_n=\infty$.
No! suppose $a_n=1$ and $x_n=\frac{1}{n^2}$ then $\sum_{n=0}^\infty a_nx_n$ is convergent and $x_n\rightarrow 0$ but $\sum_{n=0}^\infty a_n$ diverges.
