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Does there exist a sequence of real numbers $\{a_k\}_{k=1}^\infty$ such that for every $n\in\mathbb{N}$ we have $$\bigg|\sum\limits_{k=1}^na_k\bigg|\geq\sqrt{n}$$ but $$\bigg|\sum\limits_{k=1}^na^3_k\bigg|\leq \frac{c}{n}$$ for some constant $c>0$ which does not depend on $n$?

Clearly there is no such sequence of non-negative numbers, as given the second inequality above, Hölder's inequality implies $$\sum\limits_{k=1}^n|a_k\cdot 1|\leq n^{2/3}\bigg(\sum\limits_{k=1}^n|a_k|^3\bigg)^{1/3}\leq cn^{1/3}.$$

But what if not all the members of the sequence $\{a_k\}_{k=1}^\infty$ are non-negative?

Then, assuming $\bigg|\sum\limits_{k=1}^na_k\bigg|\geq\sqrt{n}$, what is the minimal possible (asymptotic) order of $\bigg|\sum\limits_{k=1}^na^3_k\bigg|$?

Is the order $O(1/n)$ achievable, and, if not, what is the smallest asymptotic order that we can have?

Aron
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  • I remembered @Calvin Lin gave a helpful comment here. Where is the comment? – River Li Feb 06 '24 at 02:03
  • Suppose that ${a_k}{k=1}^\infty$ is a required sequence. Since the series $\sum{k=1}^n a^3_k$ converges, $a_k$ tends to $0$ when $k$ tends to the infinity. Then the sign of the sum $\sum_{k=1}^n a_k$ stabilizes, because otherwise we have $|a_{n+1}|\ge \sqrt{n}+\sqrt{n+1}$ for infinitely many $n$. – Alex Ravsky Feb 06 '24 at 02:34
  • @RiverLi he deleted, he said after a certain index, all terms should have the same sign. Since each term can be bounded, one should be able to estimate the number of positive/negative terms. – Yimin Feb 06 '24 at 04:51
  • @Yimin Thanks! At that time, I think it works. – River Li Feb 06 '24 at 06:13

1 Answers1

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There exists such a sequence. Even more, we can replace $\frac{c}{n}$ by $\frac{c}{n^{1.5}}$ which is the best one can hope for since this bound implies that $|a_n| \lesssim \frac{1}{n^{0.5}}$. The construction is as follows:

$$a_{3k+1} = a_{3k+2} = 100 *(k+1)^{-0.5}, a_{3k+3} = -100*\sqrt[3]{2}*(k+1)^{-0.5}.$$

The sum of cubes of $a_{3k+1}$, $a_{3k+2}$ and $a_{3k+3}$ is zero and so the sum of the cubes of the first $n$ terms is equal to the sum of at most two last terms, so it is $O(\frac{1}{n^{1.5}})$.

As for the sum of the terms themselves, $a_{3k}+a_{3k+1}+a_{3k+2} = 100*(2-\sqrt[3]{2})*(k+1)^{-0.5}$, summing which we get something larger than $2\sqrt{n}$ if $n$ is divisible by $3$, and for other $n$ the last two terms are too small to make the difference.

I came up with it by recalling the following almost embarrasingly personal expirience: when I was a kid in the math summer camp we sometimes would be split into two teams and do some sort of team competition. If you have a rating of kids and you want to make the teams with equal number of children, you take first to the first team, second to the second team, and then alternate. If you want more balanced teams so that the sums of positions of kids are also equal, you take first and fourth to the first team, second and third to the second team, and then repeat with period $4$. To balance the sums of first $k$ powers to be equal you do the Thue--Morse sequence of length $2^{k+1}$. Now, with this idea in mind, it is not hard to make it so that some of the balance is preserved (sum of cubes in our case) while some other balance is broken, which is exactly what we did.

Aleksei Kulikov
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