Erdős Problem #268
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Erdős problems openaigpt 6 astra ultra 100% cataloguedSolvedSolvedStatement
Does the set \[ X=\left\{\left(\sum_{n\in A}\frac1n, \sum_{n\in A}\frac1{n+1},\sum_{n\in A}\frac1{n+2}\right): A\subseteq\mathbb N\hbox{ infinite},\ \sum_{n\in A}1/n<\infty\right\} \] contain a nonempty open subset of $\mathbb R^3$? Yes. Kovač first proved the three-dimensional case; Kovač and Tao subsequently proved all dimensions. We reconstruct the latter paper's local approximation and iteration in dimension three, including the approximation lemma. No novelty is claimed.
Restatement from Math-5541-A1 r1, with the source manuscript’s qualifications.
Original problem on erdosproblems.com
Statement licence: CC-BY-4.0 · Source file
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GPT 6 Astra Ultra attempt on Math-5541
Erdős Problem #268
Statement and attribution
Does the set \[ X=\left\{\left(\sum_{n\in A}\frac1n, \sum_{n\in A}\frac1{n+1},\sum_{n\in A}\frac1{n+2}\right): A\subseteq\mathbb N\hbox{ infinite},\ \sum_{n\in A}1/n<\infty\right\} \] contain a nonempty open subset of $\mathbb R^3$? Yes. Kovač first proved the three-dimensional case; Kovač and Tao subsequently proved all dimensions. We reconstruct the latter paper's local approximation and iteration in dimension three, including the approximation lemma. No novelty is claimed.Work I: coordinates with different decay rates
Define \[ f_1(x)=1/x,\quad f_2(x)=1/[x(x+1)],\quad f_3(x)=1/[x(x+1)(x+2)]. \] The invertible linear map \[ U(x,y,z)=\left(x,x-y,(x-2y+z)/2\right) \] maps $(1/n,1/(n+1),1/(n+2))$ to $(f_1(n),f_2(n),f_3(n))$. It suffices to construct an open box of sums in the latter coordinates.We prove a finite approximation lemma. For all sufficiently large integers $N$, and integers $1\le M\le N/12$, put \[ s_i=\sum_{j=1}^3f_i(jN),\qquad S(N,M)=\left\{\left(\sum_{j=1}^3f_i(jN+u_j)\right)_{i=1}^3: u_j\in[-M,M]\cap\mathbb Z\right\}. \] There are fixed $\varepsilon>0$ and $D\ge1$ such that every $y$ with $|y_i-s_i|\le\varepsilon M/N^{i+1}$ has an approximant $v\in S(N,M)$ satisfying \[ |y_i-v_i|\le D\left(N^{-i-1}+M^2N^{-i-2}\right) \quad(1\le i\le3). \tag{1} \]
Here are the details. Differentiation or Taylor expansion of these fixed rational functions gives, uniformly for the indicated integer shifts, \[ f_i(jN+u)=f_i(jN)-\frac{iu}{j^{i+1}N^{i+1}} +O\left(\frac{u^2}{N^{i+2}}\right). \tag{2} \] For nonzero integer $u$, the possible first-derivative error $O(|u|N^{-i-2})$ is absorbed by $u^2$; for $u=0$ the error is zero. More explicitly, $f_i'(x)=-ix^{-i-1}+O(x^{-i-2})$ and $f_i''(x)=O(x^{-i-2})$ on these intervals, which proves $2$. The matrix $W_{ij}=ij^{-i-1}$ is invertible: after removing nonzero row and column factors its determinant is a Vandermonde determinant at $1,1/2,1/3$. Choose $\varepsilon>0$ so that $\|W^{-1}\|_\infty\varepsilon\le1/2$. For $z_i=N^{i+1}(y_i-s_i)$, the vector $u^*=-W^{-1}z$ has coordinates in $[-M/2,M/2]$. Round them to integers $u_j$. Since $M\ge1$, they lie in $[-M,M]$, and $\|W(u-u^*)\|_\infty$ is bounded independently of $M,N$. Combining this rounding error with $2$ proves $1$.
Work II: the next block absorbs the approximation error
Let $\alpha=9/8$ and, for large integer $k$, set $N_k=\lceil\exp(\alpha^k)\rceil$, $M_k=\lfloor\sqrt{N_k}\rfloor$. Write $s_k$ for the center above, $S_k=S(N_k,M_k)$ and \[ R_k=\prod_{i=1}^3[-\varepsilon M_k/N_k^{i+1}, \varepsilon M_k/N_k^{i+1}]. \] For every sufficiently large $k$, $1$ implies \[ s_k+R_k\subseteq S_k+R_{k+1}. \tag{3} \] Indeed the error in $1$ is at most $2DN_k^{-i-1}$, whereas the next permitted error is asymptotic to $\varepsilon N_{k+1}^{-i-1/2}$. Their ratio tends to zero because \[ \alpha(i+1/2)<i+1\qquad(i=1,2,3), \] with the closest case $\alpha(7/2)=63/16<4$.Choose $k_0$ so that $3$ and all preceding conditions hold thereafter. We claim that every point of the nondegenerate box \[ \sum_{k\ge k_0}s_k+R_{k_0} \tag{4} \] is the sum of a series $\sum_{k\ge k_0}v_k$ with $v_k\in S_k$. To prove this, maintain the invariant \[ y\in\sum_{k=k_0}^{K-1}v_k+\sum_{k\ge K}s_k+R_K. \] It holds initially by $4$, and $3$ supplies $v_K$ and the next invariant. The baseline tail and the diameter of $R_K$ both tend to zero. Thus passing to the limit gives $y=\sum v_k$.
Each chosen block consists of three distinct positive integers $jN_k+u_j$. Blocks are disjoint and ordered once $k_0$ is large, since $N_{k+1}/N_k\to\infty$ and $M_k=o(N_k)$. The resulting set $A$ is infinite, and its reciprocal sum is bounded by a constant times $\sum_kN_k^{-1}<\infty$. Thus every point of $4$ belongs to $U(X)$, and the inverse image of its interior is a nonempty open subset of $X$.
Verification and outcome
Rounding is done only after solving an invertible three-variable linear system; arbitrary independent rounding of the target coordinates would not suffice. The inequality $\alpha<8/7$ ensures contraction in the fastest-decaying third coordinate. Infinite sets, distinct denominators and convergence have all been checked.SOLVED. The local approximation and infinite construction are proved above, using only elementary analysis and finite-dimensional linear algebra.