Double sum - Miklos Schweitzer 2010

Question

There is a question in the Miklos Schweitzer contest last year that keeps bugging me. Here it is:

Is there any sequence $(a_n)$ of nonnegative numbers for which $\displaystyle\sum_{n \geq 1}a_n^2 <\infty $ and $$\sum_{n \geq 1}\left(\sum_{k \geq 1}\frac{a_{kn}}{k}\right)^2=\infty\quad?$$

Answer 1

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

Answer 2

(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.

Answer 3

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.

Answer 4

I thought I had something, but as ShreevatsaR correctly remarked, I made a stupid mistake, and the below proof is false. However, Claim 1 is (I hope) correct, and perhaps the idea beyond claim 2 can be used by someone to produce a proof, so I will not delete the answer (at least for the time being).

Firstly, let $N(a) := \sum_n a_n^2, \lVert a \rVert := \sqrt{N(a)}$ and $S(a) := \sum_n (\sum_k \frac{a_{nk}}{k} )^2$ where $a$ is a sequence of nonnegative numers. We want to prove that $S(a)$ can be infinite while $N(a)$ is finite.

Claim 1

It suffices to show that the ratio $\frac{S(a)}{N(a)}$ can be made arbitrarily large (rather than actually infinite).

Proof

Suppose that the ratio $\frac{S(a)}{N(a)}$ can be made as large as we like. Then we can find, for any $M$ a sequence $a^M$ be a sequence with $N(a^M) < \frac{1}{2^{2M}}$ (so that $ \lVert a^M \rVert < \frac{1}{2^M}$) and $S(a^M) > M$. Now, define $a := \sum a^M$. It is a well defined and square-sumable sequence, since $\lVert \cdot \rVert$ is actually a norm. Now, $S(a) \geq S(a^M) > M$ for any $M$, since $a_n \geq a^M_n$ for any $n$ and all coefficients are nonnegative. Thus, we conclude that $S(a) = \infty$.

Claim 2

The ratio $\frac{S(a)}{N(a)}$ can be made arbitrarily large.

Proof

Fix an integer $M$ and consider the sequence $a_n = n[n \leq M]$ (which means $a_n = n$ for $n \leq M$ and $a_n = 0$ for $n > M$). We can compute: $$N(a) = \sum_{n=1}^M n^2 $$ and $S(a) = \sum_{n=1}^M (\sum_{k=1}^M \frac{nk}{k})^2 = \sum_{n=1}^M n^2 \cdot M^2 = M^2 N(a)$ Thus, the desired ratio is: $$\frac{S(a)}{N(a)} = M^2$$ which is obviously sufficiently large is $M$ is sufficiently large.