Double sum - Miklos Schweitzer 2010

Prof. Noam Elkies has given a answer at Mathoverflow. Here is his answer.

I figured out something of potential interest, but I am uncertain of its usefulness.

Also, I apparently can't comment yet because I don't have enough privilege points or something, so I'm "answering". (The scratchwork wouldn't fit in a comment anyway.)

Define $ \sum a_n^2 = L $ and note that $ n^2 \le \sigma_2(n) < \zeta(2) n^2 $ (this can be seen by factoring $ \sigma_2(n)n^{-2} $ and then noting that it is always a finite part of the Euler product for the Riemann zeta function but can allow arbitrarily many of its terms). Use $ (n,m) $ for greatest common divisor. Then

$$ S = \sum_{n \geq 1}\left(\sum_{k \geq 1}\frac{a_{kn}}{k}\right)^2 $$ $$ = \sum_{n,m=1}^\infty \left( \sum_{d|(n,m)} \frac{1}{(n/d)(m/d)} \right) a_n a_m $$ $$ = \sum_{n,m=1}^\infty \sigma_2((n,m)) \frac{a_n}{n} \frac{a_m}{m} $$ $$ = \sum_{n=1}^\infty \sigma_2 (n) (a_n / n)^2 + 2 \sum_{n=2}^\infty \left( \sum_{m<n} \frac{\sigma_2((n,m))}{m} a_m \right) \frac{a_n}{n} $$ $$ <\zeta(2) \left( L + 2 \sum_{n=2}^\infty \left( \sum_{m<n} \frac{(n,m)^2}{m} a_m \right) \frac{a_n}{n} \right) $$

Hence if we can prove $$ M = \sup \left\{ \frac{1}{n a_n} \sum_{m<n} \frac{(n,m)^2}{m} a_m \right\} < \infty $$ we may then establish the existence of finite upper bound

$$ S < \zeta(2) (L + 2M (L-a_1^2)) .$$

I'd look into this more (namely on how to find a lower bound practical enough for the approach from the other side) but it's the middle of the night here. Maybe later.

(This is not an answer.) The situation is even worse than Patrick Da Silva suggests. Let $a_n = \frac{b_n}{\sqrt{n}}$; then $\sum \frac{b_n^2}{n}$ converges, so $\liminf b_n = 0$. Moreover $\frac{a_{nk}}{k} = \frac{b_{nk}}{k^{3/2} \sqrt{n}}$, hence the above sum gives

$$\sum_{n \ge 1} \frac{1}{n} \left( \sum_{k \ge 1} \frac{b_{nk}}{k^{3/2}} \right)^2.$$

In particular, if $b_n$ is eventually monotonically decreasing (or even some weaker form of this assumption), then the sum in parentheses is eventually at most $b_n^2 \zeta \left( \frac{3}{2} \right)$ and so $a_n$ cannot be a counterexample to the given statement.

In other words, a counterexample needs to be fairly non-monotonic (if the statement is false). It seems like a good idea to make $a_n$ large if $n$ has many factors, but I haven't been able to do anything concrete with this.