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


1 Answer 1


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".


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