Series convergence if $\sum a_n^2 < \infty$

Question

There are quite a few simple results about convergent/divergent series derived from similar ones. Here is a question in the same spirit that I saw posted on another forum. Unfortunately, I don't have any background information on the problem:

Given a non-negative sequence $\{a_n\}_{n=1}^\infty$ such that $\sum_{n=1}^{\infty} a_n^2<\infty$, is it true that the derived series $$\sum_{n=1}^{\infty}\left(\sum_{k=1}^{\infty}\frac{a_{kn}}{k} \right)^2$$ must also converge?

At first look, the statement looks false (to me), but I have not been able to find a counter example. Maybe it's trivial but it doesn't look intuitive to me, so I would appreciate any suggestions.

Answer 1

I think the answer is no.

By functional analysis, we know that it suffices to show the existence of a sequence $a_n$ such that $\lVert a \rVert_2 = 1$ and $$\sum_{n = 1}^{\infty} \left(\sum_{j = 1}^\infty \frac{a_{jn}}{j}\right)^2$$ is arbitrarily large.

Take the first $k$ prime numbers $p_1, \cdots, p_k$, and an integer $s\ge1$. Let $\mathbb{B}_{k, s}$ denote the set $$\mathbb{B}_{k, s} = \{p_1^{i_1} p_2^{i_2} \cdots p_k^{i_k}: 0 \leq i_1, i_2, \cdots, i_k \leq s\}.$$ We consider the sequence $$a_n = \begin{cases} \frac{1}{\sqrt{|\mathbb{B}_{k, s}|}}\text{ if } n \in \mathbb{B}_{k, s}, \\ 0 \text{ otherwise}. \end{cases}.$$ Then $\lVert{a\rVert}_2 = 1$. However, if $n \in \mathbb{B}_{k, s - 1}$, then $np_i \in \mathbb{B}_{k, s}$ for each $i \in [k]$. So we have $$\sum_{j = 1}^\infty \frac{a_{jn}}{j} \geq \left(\sum_{i = 1}^k \frac{1}{p_i}\right) \cdot \frac{1}{\sqrt{|\mathbb{B}_{k, s}|}}.$$ Thus $$\sum_{n = 1}^{\infty} \left(\sum_{j = 1}^\infty \frac{a_{jn}}{j}\right)^2 \geq \left(\sum_{i = 1}^k \frac{1}{p_i}\right)^2 \frac{|\mathbb{B}_{k, s - 1}|}{|\mathbb{B}_{k, s}|} = \left(\sum_{i = 1}^k \frac{1}{p_i}\right)^2 \cdot \frac{s^k}{(s+1)^k} \gg (\log\log k)^2 \cdot \frac{s^k}{(s+1)^k}.$$ which can be arbitrarily large by tuning $s$ and $k$, as desired.