The starting point of this question is that $\sum\_{n=1}^\infty \frac{1}{n^{\alpha}} < \infty$ if and only if $\alpha > 1$.


Let $(a\_n)\_{n\in\mathbb{N}}$ be a non-negative sequence. We say that $(a\_n)$ *slowly converges* to $0$ if $\lim\_{n\to\infty} a\_n = 0$ and $$\sum\_{n=1}^\infty \frac{1}{n^{1+a\_n}} < \infty.$$


If $(a\_n), (b\_n)$ are sequences that slowly converge to $0$, does the pointwise product $(c\_n)\_{n\in\mathbb{N}}$ defined by $c\_n = a\_nb\_n$ for all $n\in\mathbb{N}$ slowly converge to $0$? (The pointwise product is the sequence $(c\_n)\_{n\in\mathbb{N}}$ defined by $c\_n = a\_nb\_n$ for all $n\in\mathbb{N}$.) 



Product of sequences "slowly" converging to $0$
The answer is no. For example $$a\_n=\frac{2\log\log n}{\log n}$$ "slowly" converges to $0$, because $n^{1+a\_n}=n\log^2n$,
but $$c\_n=\frac{\log\log n}{\log n}$$ converges to $0$ not "slowly".
As $a\_n^2<c\_n$, for large $n$, we conclude that $a\_n^2$ converges to $0$ not "slowly".


