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 $n\in\mathbb{N}$.)