Given a bounded sequence $(\beta_j)$ of positive real numbers such that \begin{align*} \sum_{j=1}^\infty\beta_j=+\infty, \end{align*} and a (strictly) monotonically decreasing sequence $(x_j)$ of positive numbers converging to 0.
Question: Is it always possible to find a sequence $(\alpha_j)$ of positive numbers, such that \begin{align*} \sum_{j=1}^\infty a_j<\infty,\text{ but }\sum_{j=n}^\infty \alpha_j\beta_j\geq x_n\text{ for infinitely many }n\in\mathbb N? \end{align*}
First note that since $(\beta_j)$ is bounded, $\sum_{j=1}^\infty\alpha_j\beta_j$ must converge if $\sum_{j=1}^\infty\alpha_j$ converges. One could of course try to set $\alpha_j=\frac{x_j-x_{j+1}}{\beta_j}$, since then $\sum_{j=n}^\infty\alpha_j\beta_j=x_n$, but $\sum_{j=1}^\infty\alpha_j$ does not have to converge. It is even worse, since it can happen that $\liminf_{j\to\infty}\alpha_j=0$ and $\limsup_{j\to\infty}\alpha_j=\infty$ hold simultaniuously in this case.
Any ideas or hints are highly appreciated. Thanks in advance!
Not always.
Let $\beta_j = \dfrac{1}{j}$ and $x_j = \dfrac{1}{1 + \log j}$. If $(\alpha_j)$ is a sequence of positive numbers with $\sum \alpha_j < +\infty$, then
$$\sum_{j = n}^{\infty} \alpha_j\beta_j \leqslant \frac{1}{n} \sum_{j = n}^{\infty} \alpha_j,$$
and the condition would imply
$$\sum_{j = n}^{\infty} \alpha_j \geqslant nx_n = \frac{n}{1 + \log n}$$
for infinitely many $n$. But that is impossible since the left hand side tends to $0$ and the right hand side tends to $+\infty$.