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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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I think $kn$ is multiplication $k\cdot n$ – Beni Bogosel Jun 1 '11 at 18:53
Now this is a good one. The case where $a_n = 1/n^{\alpha}$ shows precisely why this question is such a badass : for $\alpha > 1/2$, we have the convergence when summing $a_n^2$ but $$ \sum_{n \ge 1} \left( \sum_{k \ge 1} \frac 1{n^{\alpha} k^{\alpha+1}} \right)^2 = \sum_{n \ge 1} \frac {\zeta(\alpha+1)^2}{n^{2\alpha}} = \infty $$ because now $2 \alpha > 1$. Even worse, for the case $\alpha = 1/2$, we finally get divergence of the double sum, but now the sum of the $a_n^2$ diverges, so that looking for a sequence which satisfies this means that the sequence is "sharper" than $1/n^{\alpha}$... – Patrick Da Silva Jun 2 '11 at 5:54
I tried all kinds of examples and didn't get anywhere. – Beni Bogosel Jun 2 '11 at 8:05
Anybody else reminded of Hardy's inequality $\sum \left(\frac{a_1 + \dots + a_n}{n}\right)^2 \le 4\sum a_n^2$? – Sam Jun 4 '11 at 18:09
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. – Jakub Konieczny Jun 7 '11 at 11:40
up vote 16 down vote accepted

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

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providing a link to the original answer is a much better strategy than copying it here. For example: if Elkies edits his answer for some reason, will you update the copy? – Mariano Suárez-Alvarez Jun 12 '11 at 2:39
@Mariano: Sorry, this didn't strike my mind. Anyhow thanks for reminding :) – user9413 Jun 12 '11 at 2:49
you should thank Sir Tim, really! :) – Mariano Suárez-Alvarez Jun 12 '11 at 2:51
@Mariano: Ha ha ... :) – user9413 Jun 12 '11 at 2:58

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.

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What's $\sigma_2$? – mac Jun 10 '11 at 8:03
@mac: – joriki Jun 10 '11 at 15:54
@joriki: thanks! – mac Jun 10 '11 at 17:34

(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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By the way, here's a weaker question: can anyone even force the ratio between the desired sum and $\sum a_n^2$ to be arbitrarily large? – Qiaochu Yuan Jun 7 '11 at 0:09
I think that those tasks are equivalent. I wouldn't like to get into technical details, but once you find sequences with arbitrarily big ratio, I think you can simply multiply them by suitable coefficients, sum them up and get a sequence with the double sum infinite. – Jakub Konieczny Jun 7 '11 at 11:43
$b_n$ does not necessarily go to $0$: for example $b_n=1$ if $n=2^k$, $0$ if not. – Plop Jun 7 '11 at 15:46
Oops! Good call. Hmm. – Qiaochu Yuan Jun 7 '11 at 16:00

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).


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.


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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The inner sum should be $\sum_{k=1}^{\lfloor M/n \rfloor} \frac{nk}{k}$, not $\sum_{k=1}^M \frac{nk}{k}$. – joriki Jun 10 '11 at 15:04
Also, did you mean $N(a^M) < \frac{1}{2^{2M}}$ rather than $S(a^M) < \frac{1}{2^{2M}}$? And could you elaborate on the "since $\lVert \cdot \rVert$ is actually a norm" part? If I understand your notation $a := \sum a^M$ correctly, $a$ can't be square-summable since it is monotonically increasing. – joriki Jun 10 '11 at 15:11
Sorry for the answer full of bugs and incorrect. Obviously, I meant $N$ rather than $S$, I have corrected that one. As for the norm, I mean that the sequence of the partial sums $\sum_{M=1}^k a^M$ is a Cauchy sequence in the space of all square-sumable sequences, which follows from the triangle inequality and the fact that the sum of norms of all $a^M$ is finite. – Jakub Konieczny Jun 10 '11 at 15:17
The second part is off. As Joriki remarks in his first comment, the inner sum should be different. @Joriki: I don't get $\zeta(2)$ for the ratio, I get 3. Which should it be? – Eric Naslund Jun 10 '11 at 15:24
@Eric: You're right. I've deleted that comment; the ratio already exceeds $\zeta(2)$ for $M=4$, and is bounded by $M\cdot M^2:\sum_{n=1}^MM^2=M^3:M(M+1)(2M+1)/6<3$. – joriki Jun 11 '11 at 8:42

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