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Math-732-A2 · revision r1 · on Math-732

gpt pro 5.2 attempt on Math-732

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Unverified LLM-generated attempt. Automated screening checks form, honesty and scope, not correctness. Status at erdosproblems.com: open. Claims here do not change the source status.

Manuscript

Contents
  1. Erdős Problem #252
  2. 1) FORMAL RESTATEMENT
  3. 2) QUICK LITERATURE/CONTEXT CHECK
  4. 3) ATTACK PLAN
  5. 4) WORK
  6. 4.1 A basic lemma for factorial series
  7. 4.2 The case $k=1$ (complete proof)
  8. 4.3 The case $k=2$ (complete proof)
  9. 4.4 Why the naive bounds break for $k\ge 3$
  10. 5) VERIFICATION (adversarial check)
  11. 6) FINAL
  12. 7) COMPLETION ESTIMATE

Erdős Problem #252

1) FORMAL RESTATEMENT

Fix an integer $k\ge 1$ and define the divisor power-sum function \[ \sigma_k(n) := \sum_{d\mid n} d^k\qquad (n\in\mathbb{N}). \] Define the real number \[ \alpha_k := \sum_{n=1}^{\infty} \frac{\sigma_k(n)}{n!}. \] The question is whether $\alpha_k\notin\mathbb{Q}$ for every integer $k\ge 1$.

2) QUICK LITERATURE/CONTEXT CHECK

The irrationality of $\alpha_k$ is known unconditionally for $1\le k\le 4$: Erdős observed that $k=1,2$ are "reasonably straightforward" [Er52]; $k=3$ was proved by Schlage-Puchta [ScPu06] and independently by Friedlander--Luca--Stoiciu [FLC07]; $k=4$ was proved by Pratt [Pr22]. For general $k$, irrationality is known conditionally (Schinzel's conjecture or the prime tuples conjecture) [ScPu06,FLC07]. The problem is listed as open for general $k$ in [ErdosProblems252].

3) ATTACK PLAN

Track A (prove irrationality for each $k$): Use the standard "factorial series" method: if $\alpha_k=a/b\in\mathbb{Q}$, then for all $N\ge b$ the quantity $N!\alpha_k$ is an integer, so the tail \[ T_{k}(N):=\sum_{n>N} \frac{\sigma_k(n)}{(N+1)(N+2)\cdots n} \] must be an integer. One seeks infinitely many $N$ for which $T_k(N)$ lies strictly between two consecutive integers.

Track B (disprove): Try to exploit possible cancellation or known closed forms for related factorial series. No such mechanism is known for $\sigma_k(n)$.

Track C (bridge to known proofs): For $k\ge 3$, known proofs use deeper input (sieve + primes with controlled local factorisation). Reproducing them fully is beyond what is attempted here; instead, I give complete proofs for $k=1,2$ and explain where the naive approach fails for $k\ge 3$.

4) WORK

4.1 A basic lemma for factorial series

Lemma: Let $(a_n)_{n\ge 1}$ be integers and let $A:=\sum_{n\ge1} a_n/n!$ (assumed convergent). If $A=a/b\in\mathbb{Q}$ in lowest terms, then for every integer $N\ge b$ the tail \[ \sum_{n>N} a_n\frac{N!}{n!} \,=\, \sum_{n>N} \frac{a_n}{(N+1)(N+2)\cdots n} \] is an integer.
Proof: Fix $N\ge b$. Since $b\mid N!$, we have $N!A = N!a/b\in\mathbb{Z}$. Also \[ N!A = \sum_{n=1}^{N} a_n\frac{N!}{n!} + \sum_{n>N} a_n\frac{N!}{n!}. \] For $n\le N$ the ratio $N!/n!$ is an integer, hence the first sum is an integer. Therefore the second sum is an integer as well. ∎

4.2 The case $k=1$ (complete proof)

Write $\sigma_1(n)=\sigma(n)$. Define \[ \alpha_1 = \sum_{n\ge1} \frac{\sigma(n)}{n!}. \]
Theorem: $\alpha_1$ is irrational.
Proof: Assume for contradiction that $\alpha_1=a/b\in\mathbb{Q}$ in lowest terms. Choose a prime $p>\max\{b,100\}$ and set $N:=p-1$. Then $N\ge b$, so by Lemma Lemma the tail \[ T:=\sum_{n\ge p} \frac{\sigma(n)}{p(p+1)\cdots n} \] must be an integer.

The first term ($n=p$) satisfies $\sigma(p)=p+1$, so \[ \frac{\sigma(p)}{p} = \frac{p+1}{p} = 1 + \frac{1}{p}. \] Write \[ T = \left(1+\frac{1}{p}\right) + R,\qquad R:=\sum_{m\ge1} \frac{\sigma(p+m)}{p(p+1)\cdots (p+m)}. \] Clearly $R>0$, hence $T>1$. It remains to show $T<2$.

We use two elementary bounds:

  1. For every $n\ge1$, the divisor-counting function $\tau(n)$ satisfies $\tau(n)\le 2\sqrt{n}$ (divisors come in pairs $d$ and $n/d$, with at most $\sqrt{n}$ divisors $\le\sqrt{n}$). Hence \begin{equation}\label{wsref-0-65-71-3a-73-69-67-6d-61-5f-62-6f-75-6e-64-5f-6b-31} \sigma(n)=\sum_{d\mid n} d \le n\tau(n) \le 2n^{3/2}. \end{equation}
  2. For $m\ge 1$, \begin{equation}\label{wsref-0-65-71-3a-64-65-6e-6f-6d-5f-62-6f-75-6e-64} p(p+1)\cdots(p+m) = (p+m)\prod_{j=0}^{m-1} (p+j) \ge (p+m)\,p^m. \end{equation}
Combining (eq:sigma_bound_k1)--(eq:denom_bound), for every $m\ge1$, \[ \frac{\sigma(p+m)}{p(p+1)\cdots(p+m)} \le \frac{2(p+m)^{3/2}}{(p+m)p^m} = \frac{2\sqrt{p+m}}{p^m}. \] Therefore \[ R\le \sum_{m\ge1} \frac{2\sqrt{p+m}}{p^m}. \] Using $\sqrt{p+m}\le \sqrt{p}+\sqrt{m}$ and then $\sqrt{m}\le m$, we obtain \[ R\le 2\sqrt{p}\sum_{m\ge1}\frac{1}{p^m} + 2\sum_{m\ge1}\frac{m}{p^m} = 2\sqrt{p}\cdot \frac{1}{p-1} + 2\cdot \frac{p}{(p-1)^2}. \] Since $p>100$, we have $p-1>p/2$, hence \[ 2\sqrt{p}\cdot\frac{1}{p-1} < 2\sqrt{p}\cdot\frac{2}{p} = \frac{4}{\sqrt{p}} \le \frac{4}{10}=0.4, \] and \[ 2\cdot\frac{p}{(p-1)^2} < 2\cdot \frac{p}{(p/2)^2} = \frac{8}{p} < 0.08. \] So $R<0.48$. Consequently, \[ T = 1 + \frac{1}{p} + R < 1 + 0.01 + 0.48 < 2. \] Thus $1<T<2$, so $T$ is not an integer, contradicting Lemma Lemma. This contradiction shows $\alpha_1\notin\mathbb{Q}$. ∎

4.3 The case $k=2$ (complete proof)

Define \[ \alpha_2 = \sum_{n\ge1} \frac{\sigma_2(n)}{n!}. \]
Theorem: $\alpha_2$ is irrational.
Proof: Assume for contradiction that $\alpha_2=a/b\in\mathbb{Q}$ in lowest terms. Choose a prime $p>\max\{b,17\}$ and set $N:=p-1$. Then $N\ge b$, so by Lemma Lemma the tail \[ T:=\sum_{n\ge p} \frac{\sigma_2(n)}{p(p+1)\cdots n} \] must be an integer.

For the first term ($n=p$), since $p$ is prime, \[ \sigma_2(p)=1+p^2,\qquad \frac{\sigma_2(p)}{p} = p + \frac{1}{p}. \] Write \[ T = \left(p+\frac{1}{p}\right) + R,\qquad R:=\sum_{m\ge1} \frac{\sigma_2(p+m)}{p(p+1)\cdots (p+m)}. \] Lower bound. For $m=1$, the divisors $1$ and $p+1$ contribute, so $\sigma_2(p+1)\ge 1+(p+1)^2$, hence \[ R\ge \frac{1+(p+1)^2}{p(p+1)} = \frac{p+1}{p}+\frac{1}{p(p+1)} > 1. \] Therefore $T>p+1$.

Upper bound. We first bound the constant \[ C:=\sum_{d=1}^{\infty}\frac{1}{d^2}. \] Since $x\mapsto 1/x^2$ is decreasing on $(0,\infty)$, for every integer $d\ge 4$, \[ \frac{1}{d^2}\le \int_{d-1}^{d} \frac{dx}{x^2} = \frac{1}{d-1}-\frac{1}{d}. \] Summing from $d=4$ to $\infty$ gives $\sum_{d\ge 4} 1/d^2 \le 1/3$. Hence \begin{equation}\label{wsref-0-65-71-3a-43-5f-62-6f-75-6e-64} C = 1+\frac{1}{4}+\frac{1}{9}+\sum_{d\ge4}\frac{1}{d^2} \le 1+\frac{1}{4}+\frac{1}{9}+\frac{1}{3} = \frac{61}{36}. \end{equation} Now for any $n\ge1$, \[ \sigma_2(n)=\sum_{d\mid n} d^2 = n^2\sum_{e\mid n} \frac{1}{e^2} \le n^2\sum_{e\ge1}\frac{1}{e^2} = C n^2 \le \frac{61}{36} n^2. \] Therefore, for every $m\ge1$, \begin{equation}\label{wsref-0-65-71-3a-75-6d-5f-64-65-66} \frac{\sigma_2(p+m)}{p(p+1)\cdots(p+m)} \le \frac{61}{36}\cdot \frac{(p+m)^2}{p(p+1)\cdots(p+m)}. \end{equation} We bound $R$ by splitting off $m=1$: \[ R \le \frac{61}{36}\cdot\frac{(p+1)^2}{p(p+1)} \, +\, \frac{61}{36}\sum_{m\ge2}\frac{(p+m)^2}{p(p+1)\cdots(p+m)}. \] The first piece is \[ \frac{61}{36}\cdot\frac{(p+1)^2}{p(p+1)} = \frac{61}{36}\cdot\frac{p+1}{p}. \] For the remainder, define for $m\ge2$, \[ u_m := \frac{(p+m)^2}{p(p+1)\cdots(p+m)}. \] Then for $m\ge2$, \[ \frac{u_{m+1}}{u_m} = \frac{(p+m+1)^2}{(p+m)^2}\cdot\frac{1}{p+m+1} = \frac{p+m+1}{(p+m)^2} \le \frac{2}{p+m} \le \frac{2}{p+2}. \] Let $r:=2/(p+2)<1$. Then $u_m\le u_2\,r^{m-2}$ for all $m\ge2$, hence \[ \sum_{m\ge2} u_m \le u_2\sum_{j\ge0} r^j = \frac{u_2}{1-r}. \] Compute \[ u_2 = \frac{(p+2)^2}{p(p+1)(p+2)} = \frac{p+2}{p(p+1)}, \qquad \frac{1}{1-r} = \frac{1}{1-2/(p+2)} = \frac{p+2}{p}. \] Therefore \[ \sum_{m\ge2}u_m \le \frac{p+2}{p(p+1)}\cdot\frac{p+2}{p} = \frac{(p+2)^2}{p^2(p+1)}. \] Putting the bounds together, \[ R \le \frac{61}{36}\cdot\frac{p+1}{p} + \frac{61}{36}\cdot\frac{(p+2)^2}{p^2(p+1)}. \] Hence \begin{equation}\label{wsref-0-65-71-3a-31-6f-76-65-72-70-5f-70-6c-75-73-5f-52} \frac{1}{p}+R \le \frac{1}{p} + \frac{61}{36}\cdot\frac{p+1}{p} + \frac{61}{36}\cdot\frac{(p+2)^2}{p^2(p+1)}. \end{equation} Using $p\ge 17$, we have $(p+1)/p\le 18/17$ and $(p+2)^2/(p^2(p+1))\le (p+2)^2/p^3 \le (19/17)^2\cdot(1/17) = 361/4913$. Substituting into (eq:1overp_plus_R) gives \[ \frac{1}{p}+R \le \frac{1}{17} + \frac{61}{36}\cdot\frac{18}{17} + \frac{61}{36}\cdot\frac{361}{4913} < 2. \] Consequently $T = p + (1/p+R) < p+2$.

We have shown $p+1 < T < p+2$, so $T$ is not an integer, contradicting Lemma Lemma. This contradiction implies $\alpha_2\notin\mathbb{Q}$. ∎

4.4 Why the naive bounds break for $k\ge 3$

For $k\ge 3$, the first tail term at $N=p-1$ is \[ \frac{\sigma_k(p)}{p} = \frac{1+p^k}{p} = p^{k-1} + \frac{1}{p}, \] but the next term (at $n=p+1$) is typically of size about $(p+1)^{k-1}/p\asymp p^{k-2}$, which grows without bound. Thus bounding the entire tail in an interval of length $<1$ requires nontrivial cancellation or congruence structure beyond the elementary estimates used above. This is where the sieve-based arguments in [ScPu06,FLC07,Pr22] enter.

5) VERIFICATION (adversarial check)

  • Lemma Lemma is standard; the only needed fact is $b\mid N!$ for $N\ge b$.
  • In Theorem Theorem, the bounds $\tau(n)\le 2\sqrt{n}$ and $\sigma(n)\le n\tau(n)$ are elementary and hold for all $n$.
  • In Theorem Theorem, the inequality $\sigma_2(n)=n^2\sum_{e\mid n} 1/e^2 \le n^2\sum_{e\ge1}1/e^2$ is exact, and the bound $\sum_{e\ge1}1/e^2\le 61/36$ is justified by a telescoping integral comparison.
  • The geometric-series domination for $\sum_{m\ge2}u_m$ uses the uniform ratio bound $u_{m+1}/u_m \le 2/(p+2)<1$.

6) FINAL

UNRESOLVED (for general $k$).
  1. (i) Most advanced partial result proved here. Complete, self-contained irrationality proofs are given for $k=1$ and $k=2$ (Theorems Theorem and Theorem).
  2. (ii) Strongest obstruction / missing lemma. For $k\ge 3$, at the natural choice $N=p-1$ the second tail term (near $n=p+1$) is too large to keep the tail in an interval of length $<1$ using only crude divisor bounds; controlling this requires producing primes with special multiplicative structure in nearby integers, which is a genuinely sieve-theoretic input.
  3. (iii) Next concrete steps. To extend beyond $k=2$ by elementary means, one would need a systematic way to select $N$ so that the early tail terms have nearly-integer sum and the remaining tail is $<1$; this is exactly what the known arguments for $k=3,4$ accomplish via deep distribution results for primes and almost-primes.
  4. (iv) Counterexample search description. A counterexample would be an explicit $k$ and rational value of $\alpha_k$. Since $\sigma_k(n)\in\mathbb{Z}$, one can compute high-precision approximations of $\alpha_k$ and attempt rational reconstruction for bounded denominators; this cannot certify irrationality, but can falsify specific rational hypotheses.

7) COMPLETION ESTIMATE

$\mathbf{97\%}$ confidence in the correctness of the complete proofs for $k=1,2$ and the cited status "known for $k\le 4$; open for general $k$".