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.

@ChristianRemling Ha! That's why I hate when people write $a_{ij}$ instead of $a_{i,j}$ for the entries of a matrix, for instance. In this case, the OP did not abuse notation. The number $a_{kn}$ is the $(kn)^{\text{th}}$ element of the sequence $(a_n)_n$. — mathworker21
– mathworker21, Commented Nov 29, 2023 at 23:24
@Christian Remling: presumably $kn$ in $a_{kn}$ is the product of $k$ and $n$. — Alexandre Eremenko
– Alexandre Eremenko, Commented Nov 29, 2023 at 23:25
If the sequence is submultiplicative, $a_{kn}\le a_ka_n$, then by Cauchy-Schwarz the derived series is bounded in terms of the first one (maybe this gives a hint for a counterexample) — Pietro Majer
– Pietro Majer, Commented Nov 30, 2023 at 0:00

Iosif Pinelis · Accepted Answer · 2023-11-30 03:56:41Z

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.

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

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