How prove exists a sequence $\{a_{n}\}$ of real numbers such that $\sum_{n=1}^{\infty}a^2_{n}<\infty,\sum_{n=1}^{\infty}|a_{n}b_{n}|=\infty$ Suppose that the series $\displaystyle\sum_{n=1}^{\infty}b^2_{n}$ of postive numbers diverges. Prove that
 there exists a sequence $\{a_{n}\}$ of real numbers such that
$$
\sum_{n=1}^{\infty}a^2_{n}<\infty
\quad\text{and}\quad
\sum_{n=1}^{\infty}|a_{n}b_{n}|=\infty.
$$
My try: maybe this Cauchy-Schwarz inequality have usefull
$$\Big(\sum_{n=1}^{\infty}a^2_{n}\Big)\Big(\sum_{n=1}^{\infty}b^2_{n}\Big)\ge
\Big(\sum_{n=1}^{\infty}a_{n}b_{n}\Big)^2$$

Define
$$
s_n=\sum_{k=1}^nb_k^2
$$
and
$$
a_n=\frac{b_n}{\sqrt{s_ns_{n-1}}}
$$
Without loss of generality, assume $b_1\ne0$.

Since $u-1\ge\log(u)$ for $u\gt0$,
$$
\begin{align}
\sum_{k=2}^n a_kb_k
&=\sum_{k=2}^n\frac{s_k-s_{k-1}}{\sqrt{s_ks_{k-1}}}\\
&=\sum_{k=2}^n\sqrt{\frac{s_k}{s_{k-1}}}-\sqrt{\frac{s_{k-1}}{s_k}}\\
&\ge\sum_{k=2}^n\sqrt{\frac{s_k}{s_{k-1}}}-1\\
&\ge\frac12\sum_{k=2}^n\log\left(\frac{s_k}{s_{k-1}}\right)\\[8pt]
&=\frac12(\log(s_n)-\log(s_1))
\end{align}
$$
Therefore,
$$
\sum_{k=1}^\infty a_kb_k=\infty
$$

$$
\begin{align}
\sum_{k=2}^n a_k^2
&=\sum_{k=2}^n\frac{s_k-s_{k-1}}{s_ks_{k-1}}\\
&=\sum_{k=2}^n\frac1{s_{k-1}}-\frac1{s_k}\\
&=\frac1{s_1}-\frac1{s_n}
\end{align}
$$
Therefore,
$$
\sum_{k=2}^\infty a_k^2=\frac1{b_1^2}
$$
