A large set is a set $A$ of positive integers such that $$\sum_{a\in A}\frac{1}{a}$$ is a divergent series. Is set $A\subset\mathbb{N}$ that contains arbitrarily long arithmetic progressions a large set?
2026-03-26 17:42:19.1774546939
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Example of a large set
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No, because we can take long arithmetic progressions using arbitrarily large numbers.
Consider, say, the set $$A=(1)\cup (8,9)\cup (27, 28, 29)\cup (64,65,66,67)\cdots$$
To be precise we consider, for each $n$, the $n$ consecutive natural numbers starting at $n^3$. Calling that progression $A_n$, it is clear that $\sum_{a\in A_n} \frac 1a<\frac 1{n^2}$ (at least for $n>1$, for $n=1$ we have equality). so we see, by comparison with $\sum \frac 1{n^2}$ that the sum taken over the union of the $A_n$ converges.
No. Take$$A=\left\{1,2^2,2^2+1,2^4,2^4+1,2^4+2,2^4+3,\ldots\right\}.$$It contains arbitrarily long arithmetic progressions, but it is not a large set.