Does $\sum\limits_{i=1}^{\infty}|a_i||x_i| < \infty$ whenever $\sum\limits_{i=1}^{\infty} |x_i| < \infty $ imply $(a_i)$ is bounded?

If $a_n$ is unbounded, then there exist integers $0 < n_1 < n_2 < \cdots \to \infty$ such that $|a_{n_k}| > k^2.$ Define $x_n$ as follows: $x_{n_k} = 1/k^2, k = 1,2, \dots,$ $x_n=0$ for all other $n.$ Then $\sum |x_n| < \infty,$ while $\sum |a_n||x_n|$ has infinitely many terms $> 1,$ hence diverges, contradiction.

Hint: Look at this simple fact: If the positive series $\sum a_n$ diverges and $s_n=\sum\limits_{k\leqslant n}a_k$ then $\sum \frac{a_n}{s_n}$ diverges as well.

Hint2:

Consider $x_n = \dfrac{1}{\sum\limits_{i=1}^{n}a_i}$