TITLE: Absolute summability, square-summability and $\sum_t(\log t)^2 x_t^2<\infty$, which is the strongest?
QUESTION [5 upvotes]: I'm working through a few convergence conditions for time series. In particular, I have run into
$$
\sum_{t=1}^\infty|x_t|<\infty\tag{$*$},
$$
$$
\sum_{t=1}^\infty x_t^2<\infty,\tag{$**$}
$$
$$
\sum_{t=1}^\infty(\log t)^2x_t^2<\infty.\tag{$***$}
$$
In all series above, $\{x_t\}$ denotes a sequence of real numbers. I have already proved that absolute summability is a strictly stronger condition than square-summability. I have also proved that ($***$) implies ($**$). But how do I compare ($*$) and ($***$) please? Could you also please give a hand on the reverse implication ($**$) $\implies$ ($***$)? I suspect it doesn't hold but I don't have a counter example.
Thank you very much.

REPLY [3 votes]: (*) does not imply (***): to see this, set $$x_t=\left\{\begin{array}{c l} 1/n^2 , & t=2^{n^2}, n\in\mathbb N\\ 0, & {\rm otherwise}\end{array}\right.$$ The nonzero terms of $(x_t)$ are the terms $1/n^2$ for $n\in\mathbb N$, so $\displaystyle\sum_{t=1}^{\infty}|x_t|$ converges. But, the series $\displaystyle\sum_{t=1}^{\infty}\log^2t\cdot x_t^2$ does not converge, since $$\log^2(2^{n^2})\cdot x_{2^{n^2}}^2=n^4\log^22\cdot\frac{1}{n^4}=\log^22,$$ which shows that the sequence $\log^2t\cdot x_t^2$ does not go to $0$ as $t\to\infty$.