I have the following question, whose inverse question can be done by the well-known Cauchy–Schwarz inequality. But I do not know how to solve this question:
Suppose that $\{a_n\}_{n=1}^\infty$ is a sequence of real numbers such that \begin{equation} \sum_{n=1}^\infty a_n b_n \quad \text{is a convergent series whenever}\quad\sum_{n=1}^\infty b_n^2<\infty. \end{equation} Show that $\sum_{n=1}^\infty a_n^2<\infty$.
Proof by contraposition. Suppose $\sum a_n^2 = \infty$. Using the Cauchy criterion, show that there is a positive number $C$ and a sequence $\{F_k\}$ of disjoint finite subsets of $\Bbb N$ such that $$\sum_{n\in F_k} a_n^2 > C^2\quad (k = 1,2,3,\ldots)$$ Set $s_k = \sum\limits_{n\in F_k} a_n^2$, for $k = 1,2,3,\ldots$. Define $\{b_n\}$ by setting $b_n = \dfrac{a_n}{k\sqrt{s_k}}$ for $n\in F_k$, and $b_n = 0$ if $n$ is not in any of the $F_k$. The sum $$\sum b_n^2 = \sum_k \frac{1}{k^2}\sum_{n\in F_k} \frac{a_n^2}{s_k} = \sum_k \frac{1}{k^2} < \infty$$ and the sum $$\sum a_nb_n = \sum_k \frac{1}{k}\sum_{n\in F_k} \frac{a_n^2}{\sqrt{s_k}} > \sum_k \frac{C}{k} = \infty$$
