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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 $\sum_{n \geq 1}\left(\sum_{k \geq 1}\frac{a_{kn}}{k}\right)^2=\infty\quad?$

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    Another sad fact: not only cannot $a_n$ be of the form $n^{-\alpha}$ but also this sequence cannot be multiplicative in the sence $a_{nm} = a_m a_n$. If it was multiplicative, the double sum would equal to $\sum_n a_n^2 \sum_k \frac{a_k}{k}$. The first term is finite from assumption, for the second one, apply Cauchy-Schwarz.2011-06-07

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

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    @Mariano: Ha ha ... :)2011-06-12
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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 $ <\zeta(2) \left( L + 2 \sum_{n=2}^\infty \left( \sum_{m

Hence if we can prove $ M = \sup \left\{ \frac{1}{n a_n} \sum_{m 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.

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    @mac: http://en.wikipedia.org/wiki/Divisor_function2011-06-10
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(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.

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    Oops! Good call. Hmm.2011-06-07
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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.

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    Hey, I didn't remark anything. I don't know why you're giving me credit. :-)2011-06-12