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Math-736-A2 · revision r1 · on Math-736

gpt pro 5.2 attempt on Math-736

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Contents
  1. Erdős Problem #264
  2. 1) FORMAL RESTATEMENT
  3. 2) QUICK LITERATURE/CONTEXT CHECK (only if browsing available)
  4. 3) ATTACK PLAN
  5. 4) WORK (with full details)
  6. 4.1. A concrete counterexample for $a_n=2^n$ (bounded perturbations can give a rational sum)
  7. 4.2. The case $a_n=n!$
  8. 5) VERIFICATION / SANITY CHECK
  9. 6) FINAL
  10. 7) COMPLETION ESTIMATE

Erdős Problem #264

1) FORMAL RESTATEMENT

Definition (Type 3 irrationality sequence): A sequence of integers $(a_n)_{n\ge 1}$ is a Type 3 irrationality sequence if for every bounded sequence of integers $(b_n)_{n\ge 1}$ satisfying \[ b_n \neq 0 \ \text{and}\ a_n+b_n\neq 0 \quad\text{for all }n, \] the series $\sum_{n\ge 1} \frac{1}{a_n+b_n}$ converges and its sum is an irrational real number.

The question asks whether the specific choices $a_n=2^n$ and $a_n=n!$ are Type 3 irrationality sequences.

2) QUICK LITERATURE/CONTEXT CHECK (only if browsing available)

  • Kovač--Tao $2024/2025$ introduced a quantitative criterion implying a sequence is not Type 3; in particular they show $a_n=2^n$ fails to be Type 3 (indeed, there exists bounded $b_n$ for which the sum is rational). They also show any strictly increasing $a_n$ with $\limsup a_{n+1}/a_n<\infty$ is not Type 3, so "pure exponential" growth cannot work.
  • In the opposite direction, Kovač--Tao prove that for any function $F$ with $F(n+1)/F(n)\to\infty$ there exists a Type 3 irrationality sequence with $a_n\sim F(n)$; in particular there exist Type 3 sequences asymptotic to $n!$, but this does not decide whether the exact sequence $a_n=n!$ itself is Type 3.

3) ATTACK PLAN

  1. Disprove the Type 3 property for $a_n=2^n$ by constructing (or at least proving existence of) a bounded integer sequence $b_n$ such that \[\sum_{n\ge 1}\frac{1}{2^n+b_n}\in\mathbb{Q}. \] It suffices to produce one rational value.
  2. Show the existence of such a $b_n$ by an "overlapping-interval"/digit-expansion argument: allow $b_n\in\{1,2,3,4,5\}$, prove the set of achievable sums is an interval, then pick a rational point inside that interval.
  3. For $a_n=n!$, summarize what is known and why the same interval method fails for fixed bounded $b_n$ (the gaps between $1/(n!+b)$ values dwarf the tail width). Record the open part.

4) WORK (with full details)

4.1. A concrete counterexample for $a_n=2^n$ (bounded perturbations can give a rational sum)

Fix $B=5$. For $n\ge 1$ define the "tail bounds" \[ L_n := \sum_{k=n}^{\infty}\frac{1}{2^k+B},\qquad U_n := \sum_{k=n}^{\infty}\frac{1}{2^k+1}, \] so $0<L_n\le U_n<\infty$ and $L_n,U_n\to 0$ as $n\to\infty$.

For each $n\ge 1$ and each $b\in\{1,2,3,4,5\}$ set \[ I_{n}(b):=\left[\frac{1}{2^n+b}+L_{n+1},\ \frac{1}{2^n+b}+U_{n+1}\right]. \] Intuitively, if we choose the $n$th "digit" $b_n=b$, then the remaining tail sum lies in $[L_{n+1},U_{n+1}]$, so the total sum lies in $I_n(b)$.

Lemma (Overlap of digit intervals for $B=5$): For every $n\ge 1$ the union $\bigcup_{b=1}^{5} I_n(b)$ is exactly the interval $[L_n,U_n]$.
Proof: First note the extreme endpoints match: \[ \min I_n(5)=\frac{1}{2^n+5}+L_{n+1}=L_n, \qquad \max I_n(1)=\frac{1}{2^n+1}+U_{n+1}=U_n. \] Thus $\bigcup_{b=1}^{5} I_n(b)\subseteq [L_n,U_n]$ and the union reaches both endpoints.

It remains to show that consecutive intervals overlap, i.e. \begin{equation}\label{wsref-0-65-71-3a-6f-76-65-72-6c-61-70-5f-63-6f-6e-64-69-74-69-6f-6e} I_n(b)\cap I_n(b+1)\neq\varnothing\quad\text{for }b=1,2,3,4. \end{equation} Since $\frac{1}{2^n+b}$ decreases in $b$, the intervals are ordered from top to bottom, and (eq:overlap_condition) is equivalent to \[ \frac{1}{2^n+(b+1)}+U_{n+1}\ge \frac{1}{2^n+b}+L_{n+1} \iff U_{n+1}-L_{n+1}\ge \left(\frac{1}{2^n+b}-\frac{1}{2^n+b+1}\right). \] The right-hand side is maximized at $b=1$, so it suffices to show \begin{equation}\label{wsref-0-65-71-3a-6b-65-79-5f-69-6e-65-71-5f-74-61-72-67-65-74} U_{n+1}-L_{n+1}\ \ge\ \frac{1}{(2^n+1)(2^n+2)}. \end{equation} Compute the tail width explicitly: \[ U_{n+1}-L_{n+1} =\sum_{k=n+1}^{\infty}\left(\frac{1}{2^k+1}-\frac{1}{2^k+5}\right) =\sum_{k=n+1}^{\infty}\frac{4}{(2^k+1)(2^k+5)}. \] Fix $n$ and write $x:=2^{-(n+1)}\in(0,1/4]$. For every $k\ge n+1$ we have $2^{-k}\le x$, hence \[ (2^k+1)(2^k+5) =2^{2k}(1+2^{-k})(1+5\cdot 2^{-k}) \le 2^{2k}(1+x)(1+5x). \] Therefore \[ \frac{4}{(2^k+1)(2^k+5)}\ge \frac{4}{2^{2k}(1+x)(1+5x)}=\frac{4^{1-k}}{(1+x)(1+5x)}. \] Summing this lower bound gives \begin{equation}\label{wsref-0-65-71-3a-74-61-69-6c-77-69-64-74-68-5f-6c-62} U_{n+1}-L_{n+1} \ge \frac{1}{(1+x)(1+5x)}\sum_{k=n+1}^{\infty}4^{1-k} =\frac{1}{(1+x)(1+5x)}\cdot \frac{4}{3}\cdot 4^{-n}. \end{equation} On the other hand, \[ \frac{1}{(2^n+1)(2^n+2)} =\frac{1}{2^{2n}(1+2^{-n})(1+2\cdot 2^{-n})} =\frac{4^{-n}}{(1+2x)(1+4x)} \qquad\left(\text{since }2^{-n}=2x\right). \] Thus (eq:key_ineq_target) will follow from \[ \frac{4}{3}\cdot\frac{1}{(1+x)(1+5x)}\ \ge\ \frac{1}{(1+2x)(1+4x)}. \] Clearing denominators (all positive) reduces this to \[ 4(1+2x)(1+4x)\ \ge\ 3(1+x)(1+5x). \] Expanding both sides yields \[ 4(1+6x+8x^2)\ge 3(1+6x+5x^2) \iff 1+6x+17x^2\ge 0, \] which is true for all $x\ge 0$. This proves (eq:key_ineq_target), hence all overlaps (eq:overlap_condition). Therefore the union $\bigcup_{b=1}^{5} I_n(b)$ is connected and spans from $L_n$ to $U_n$, i.e. it equals $[L_n,U_n]$. ∎

Lemma (Digit-by-digit construction): Fix any real number $x\in[L_1,U_1]$. Then there exists a sequence $(b_n)_{n\ge 1}$ with each $b_n\in\{1,2,3,4,5\}$ such that \[ x=\sum_{n=1}^{\infty}\frac{1}{2^n+b_n}. \]
Proof: We construct $b_n$ recursively. Set $r_1:=x$.

Assume $r_n\in[L_n,U_n]$ is defined. By the previous lemma, \[ [L_n,U_n]=\bigcup_{b=1}^{5} I_n(b), \] so there exists at least one $b_n\in\{1,2,3,4,5\}$ such that $r_n\in I_n(b_n)$. Choose one such $b_n$ (for definiteness, the smallest such $b_n$). Define \[ r_{n+1}:=r_n-\frac{1}{2^n+b_n}. \] Because $r_n\in I_n(b_n)$, we have $r_{n+1}\in[L_{n+1},U_{n+1}]$ by the definition of $I_n(b_n)$. This closes the induction.

Iterating the relation $r_{n}=\frac{1}{2^n+b_n}+r_{n+1}$ gives for every $N\ge 1$, \[ x=r_1=\sum_{n=1}^{N}\frac{1}{2^n+b_n}+r_{N+1}. \] Finally $0\le r_{N+1}\le U_{N+1}\to 0$ as $N\to\infty$, so passing to the limit yields $x=\sum_{n\ge 1}\frac{1}{2^n+b_n}$. ∎

We now exhibit a rational $x\in[L_1,U_1]$.

Lemma ($\tfrac{3}{4}$ lies in the digit interval): One has $\frac{3}{4}\in[L_1,U_1]$.
Proof: Lower endpoint: since $2^n+5\ge 2^n$ for $n\ge 2$, \[ L_1=\sum_{n=1}^{\infty}\frac{1}{2^n+5} \le \frac{1}{7}+\sum_{n=2}^{\infty}\frac{1}{2^n} =\frac{1}{7}+\frac{1}{2}=\frac{9}{14}<\frac{3}{4}. \]

Upper endpoint: clearly $U_1$ dominates any partial sum. Direct computation gives \[ \sum_{n=1}^{7}\frac{1}{2^n+1} =\frac{3559541}{4703985} =\frac{3}{4}+\frac{126209}{18815940} >\frac{3}{4}. \] Therefore $U_1>\frac{3}{4}$. Combining both inequalities yields $\frac{3}{4}\in[L_1,U_1]$. ∎

Proposition (Disproof for $a_n=2^n$): Let $a_n=2^n$. Then $(a_n)$ is not a Type 3 irrationality sequence.
Proof: By the previous lemma, $x=\tfrac{3}{4}$ lies in $[L_1,U_1]$. By the digit-by-digit construction lemma, there exists a bounded integer sequence $b_n\in\{1,2,3,4,5\}$ such that \[ \sum_{n=1}^{\infty}\frac{1}{2^n+b_n}=\frac{3}{4}\in\mathbb{Q}. \] This is a direct violation of the Type 3 definition for $a_n=2^n$. ∎
Remark (An explicit initial segment): Running the "choose the smallest admissible digit" rule produces, for example, \[ (b_n)_{n=1}^{40}=1,1,1,5,4,3,5,5,5,3,5,2,3,2,5,2,4,4,3,2,3,5,4,5,3,3,3,5,2,2,2,2,2,2,2,3,2,4,4,3,\ldots \] which yields partial sums rapidly converging to $3/4$.

4.2. The case $a_n=n!$

The above interval-overlap method is tailored to exponential denominators: the "digit gaps" \[ \frac{1}{2^n+b}-\frac{1}{2^n+(b+1)}\asymp 4^{-n} \] are of the same order as the tail flexibility $U_{n+1}-L_{n+1}\asymp 4^{-n}$.

For $a_n=n!$, if we again restricted to a fixed bounded range $b\in\{1,\dots,B\}$, then the digit gaps are on the order of $1/(n!)^2$, but the tail flexibility is on the order of \[ \sum_{k\ge n+1}\frac{1}{(k!)^2}\asymp \frac{1}{((n+1)!)^2}=\frac{1}{(n+1)^2}\cdot\frac{1}{(n!)^2}, \] which is smaller by a factor $(n+1)^{-2}$. Thus no fixed $B$ can yield the kind of uniform overlap we used above.

At present, the exact Type 3 status of $a_n=n!$ is not settled by the argument above. What is known (from the general existence theorem mentioned in the literature check) is that one can construct Type 3 sequences $(a_n)$ with $a_n\sim n!$ (even with $a_n=n!+o(n!)$), but this does not imply that the unperturbed factorial sequence itself has the Type 3 property.

5) VERIFICATION / SANITY CHECK

  • All series considered with denominators $2^n+b_n$ and bounded $b_n$ converge absolutely, since $2^n+b_n\asymp 2^n$.
  • The constructed $b_n$ satisfy the constraints $b_n\neq 0$ and $2^n+b_n\neq 0$ automatically (here $b_n\in\{1,2,3,4,5\}$).
  • The overlap inequality was reduced to an explicit polynomial inequality $1+6x+17x^2\ge 0$, so no hidden analytic assumptions remain.

6) FINAL

UNRESOLVED.

  • (i) What is proved here: $a_n=2^n$ is not Type 3, by exhibiting a bounded choice of $b_n\in\{1,\dots,5\}$ for which $\sum_{n\ge 1}\frac{1}{2^n+b_n}=\tfrac{3}{4}\in\mathbb{Q}$.
  • (ii) What remains open: whether $a_n=n!$ is Type 3 (i.e., whether $\sum\frac{1}{n!+b_n}$ is irrational for every bounded integer perturbation $b_n\neq 0$ with $n!+b_n\neq 0$).
  • (iii) Most plausible next step: develop methods that exploit the arithmetic structure of $n!$ (e.g., strong divisibility and factorial-base expansions) to force irrationality for all bounded perturbations; the interval-overlap method above cannot work uniformly for fixed bounded $b_n$.
  • (iv) Completion estimate: resolving the factorial case likely requires a new invariant or approximation argument; at minimum, one would need either (a) a universal irrationality criterion that applies to all bounded perturbations of $n!$, or (b) an explicit bounded perturbation producing a rational sum.

7) COMPLETION ESTIMATE

For the remaining $a_n=n!$ case, one would need to either:

  1. produce an explicit bounded $b_n$ with $\sum 1/(n!+b_n)\in\mathbb{Q}$ (a counterexample), or
  2. prove a uniform irrationality criterion that works against every bounded perturbation of $n!$.
Either direction appears to require substantially new ideas beyond the overlap/digit method used for $2^n$.