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?

Question

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Suppose $(a_i)$ is a sequence in $\Bbb R$ such that $\sum\limits_{i=1}^{ \infty} |a_i||x_i| < \infty$ whenever $\sum\limits_{i=1}^{\infty} |x_i| < \infty$. Then is $(a_i)$ a bounded sequence?

Look at the end of the question for the right answer.

If the statement '$(a_i)$ is a properly divergent sequence implies that there exists some $k \in \Bbb N$ such that $\sum\limits_{i=1}^{\infty} {1/{a_i}}^k$ is convergent' was true, we could have easily proven $(a_i)$ is bounded by using sub-sequences but since that is dis-proven by $ln(n)$, can we use something around it? Like can all the functions which do not satisfy the 'statement' I mentioned be considered as a special case of functions?

Correct Answer - Yes, $(a_i)$ is bounded.

Source - Tata Institute of Fundamental Research Graduate Studies 2013

Answer 1

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.

Answer 2

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