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Proposition: Suppose $A\subset \mathbb{N}$ is a large set in the sense that

$$ \sum_{n\in A} \frac{1}{n} = \infty.$$

Then there exists $a_1 < a_2 < \ldots < a_k,\ $ (not necessarily consecutive) members of $A,$ such that

$$ a_k - a_{k-1} \leq a_{k-1} - a_{k-2} \leq \ldots \leq a_2 - a_1 $$

for every $k\in\mathbb{N}.$

In other words, for every $k,$ there is a subsequence of $A$ of length $k$ such that the differences (of consecutive members of the subsequence) are decreasing, also known as, the subsequence is concave.

From the ratio test, we have that, if $(x_n)_{n\in\mathbb{N} } $ is $A$ in (increasing) sequence form, then $\limsup_{n\to\infty} \left\lvert \frac{x_{n+1}}{x_n} \right\rvert =1. $ Maybe this can be used to answer the question, but I don't see how straight away. But maybe it is a helpful fact to bear in mind.

My first thought to tackle this problem was to look at Erdos et al's results in bof's answer to my earlier question here. However, that question and bof's answer is about sequences with no concave and no convex sub-sequences, whereas this question is only about concave sub-sequences, or if we're trying to prove the contrapositive: sequences with no concave subsequence of length $k$, but are allowed to have convex sub-sequences of arbitrary length..

I wonder if we can show that sequences that do not contain concave sequences of arbitrary length are quasi-exponential/geometric in some sense? That would prove the contrapositive of the proposition...

Adam Rubinson
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1 Answers1

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The affirmative answer to a previous question of mine means that for every $k$ no matter how large, $A$ will contain a quasi-arithmetic progression of length $k.$

Now note that every quasi-A.P. of length $k$ has a concave subsequence of length at least $\ \log_2 k - 1.\ $ For example, using the above theorem, we can find a quasi-arithmetic progression of $A$ of length $k=64.$ That is, there exists intervals $\ I_0,\ I_1,\ I_2,\ I_3,\ \ldots,\ I_{63}$ of equal width, such that $\ a_0\in I_0,\ a_1\in I_1,\ a_2\in I_2,\ a_3\in I_3,\ \ldots,\ a_{63} \in I_{63}.$ Then $a_0,\ a_{32},\ a_{32+16},\ a_{32+16+8},\ a_{32+16+8+4},\ a_{32+16+8+4+2}\ $ is a concave subsequence of $A,\ $ and we are done, since we can do all this for arbitrarily large $k,$ and $\ \log_2 k - 1\to\infty\ $ as $\ k\to\infty.$

Adam Rubinson
  • 24,300