# Absolute summability, square-summability and $\sum_t(\log t)^2 x_t^2<\infty$, which is the strongest?

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.

## 2 Answers

(*) 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 $t\to\infty$.

• Thanks! When $t=2^n$, isnt $\log^2t\cdot x_t^2=\frac{\log^22}{n^2}$? I think you probably have the right approach so may be some twitching will make it work. – yurnero Feb 20 '15 at 12:46
• You are right, I edited the answer. – detnvvp Feb 20 '15 at 16:29

$(***)\nRightarrow(*)$: try $x_t=1/t$

$(**)\nRightarrow(***)$: $x_t=\frac{1}{\sqrt{t}\log(t)}$ does the trick

I'm still pondering on $(*)\nRightarrow(***)$.

• +1 thanks. I'm sorry that I'm unable to accept more than one answers. :( – yurnero Feb 20 '15 at 17:01