gpt pro 5.4 attempt on Math-5533

Erdős Problems #256--#258: Round 2 continuation

Erdős Problem #258

1) ROUND-2 OBJECTIVE

I am again using the fail-safe strict-progress option. The full problem remains open, but Round 2 will prove a broad new sufficient condition that is not covered by the known monotone case:

If $a_n\to\infty$ and \[ \frac{\tau(n)}{a_n}\to 0, \] then \[ \sum_{n=1}^{\infty}\frac{\tau(n)}{a_1a_2\cdots a_n} \] is irrational.

This is the most promising direction because Round 1 showed that monotonicity is the hypothesis used by Erdős--Straus, but it did not identify a replacement mechanism. The new mechanism is a scaled-tail/fractional-part argument.

2) Round-1 FOUNDATION USED

I use the following Round-1 material.

1. The convergence of \[ T(a_1,a_2,\dots):=\sum_{n=1}^{\infty}\frac{\tau(n)}{a_1a_2\cdots a_n} \] for every integer sequence with $a_n\to\infty$.
2. The known Erdős--Straus theorem: if $(a_n)$ is monotone nondecreasing and $a_n\to\infty$, then $T(a_1,a_2,\dots)$ is irrational.
3. The Round-1 gap: without monotonicity, there was no replacement for the mixed-radix control used in the classical proof.

3) NEW INSIGHT / TOOL (ROUND-2)

The new tool is the following scaled-tail criterion.

Scaled-tail criterion. Let \[ Q_n:=a_1a_2\cdots a_n, \qquad T:=\sum_{n=1}^{\infty}\frac{\tau(n)}{Q_n}, \qquad R_n:=\sum_{m=n+1}^{\infty}\frac{\tau(m)}{a_{n+1}a_{n+2}\cdots a_m}. \] If $R_n\to 0$, then $T\notin\mathbb{Q}$.

This is genuinely new in the present investigation. It replaces monotonicity by a direct control of the tail after multiplying by $Q_n$.

4) ATTACK PLAN (ROUND-2)

The remaining Round-1 obstacle was: after multiplying by $Q_n$, how can one force a contradiction from rationality without assuming monotonicity? The plan is:

1. compute the fractional part of $Q_nT$;
2. show it equals the scaled tail $R_n$;
3. if $R_n\to 0$, compare this fractional part with the discrete set of possible fractional parts of a rational number.

Then I will prove $R_n\to 0$ under the concrete hypothesis $\tau(n)/a_n\to 0$.

5) WORK (ROUND-2)

Set \[ Q_n:=a_1a_2\cdots a_n, \qquad T:=\sum_{n=1}^{\infty}\frac{\tau(n)}{Q_n}. \] For each $n\ge 1$, define the scaled tail \[ R_n:=\sum_{m=n+1}^{\infty}\frac{\tau(m)}{a_{n+1}a_{n+2}\cdots a_m}. \] Because all summands are positive, we have $R_n>0$ for every $n$.

Lemma 1 (fractional-part identity).

For every $n\ge 1$, \[ Q_nT=\sum_{m=1}^{n}\tau(m)\frac{Q_n}{Q_m}+R_n, \] and hence, whenever $R_n<1$, \[ \{Q_nT\}=R_n, \] where $\{x\}$ denotes the fractional part of $x$.

Proof. Multiply the series defining $T$ by $Q_n$: \[ Q_nT =\sum_{m=1}^{n}\tau(m)\frac{Q_n}{Q_m}+\sum_{m=n+1}^{\infty}\tau(m)\frac{Q_n}{Q_m}. \] Since $Q_m=Q_n(a_{n+1}\cdots a_m)$ for $m>n$, the second sum is exactly $R_n$. For $m\le n$, the quotient $Q_n/Q_m=a_{m+1}\cdots a_n$ is an integer, so the first sum is an integer. Therefore, if $0<R_n<1$, the fractional part of $Q_nT$ is $R_n$. $\square$

Lemma 2 (scaled-tail criterion).

If $R_n\to 0$, then $T\notin\mathbb{Q}$.

Proof. Assume for contradiction that $T=p/r\in\mathbb{Q}$ in lowest terms with $r\ge 1$. For each $n$, the number $Q_nT$ is rational with denominator dividing $r$, so its fractional part belongs to the finite set \[ \left\{0,\frac1r,\frac2r,\dots,\frac{r-1}{r}\right\}. \] Since $R_n\to 0$ and $R_n>0$, there exists $N$ such that \[ 0<R_n<\frac1r\qquad (n\ge N). \] By Lemma 1, for all sufficiently large $n$ we have $\{Q_nT\}=R_n$. But no element of $ \{0,1/r,2/r,\dots,(r-1)/r\} $ lies strictly between $0$ and $1/r$. This contradiction proves $T\notin\mathbb{Q}$. $\square$

Lemma 3 (a concrete sufficient condition).

If \[ \frac{\tau(n)}{a_n}\to 0, \] then $R_n\to 0$.

Proof. Fix $\varepsilon>0$. Since $a_n\to\infty$, there exists $N_1$ such that $a_n\ge 2$ for all $n\ge N_1$. Since $\tau(n)/a_n\to 0$, there exists $N_2$ such that \[ \frac{\tau(n)}{a_n}\le \varepsilon\qquad (n\ge N_2). \] Let $N:=\max\{N_1,N_2\}$. For $n\ge N$ we get \[ R_n =\sum_{m=n+1}^{\infty}\frac{\tau(m)/a_m}{a_{n+1}a_{n+2}\cdots a_{m-1}} \le \varepsilon\sum_{m=n+1}^{\infty}\frac{1}{a_{n+1}a_{n+2}\cdots a_{m-1}}. \] For the term $m=n+1$ the denominator is the empty product, equal to $1$. For $m=n+1+j$ with $j\ge 1$, there are $j$ factors, each at least $2$, so \[ \frac{1}{a_{n+1}a_{n+2}\cdots a_{m-1}}\le 2^{-j}. \] Hence \[ R_n\le \varepsilon\sum_{j=0}^{\infty}2^{-j}=2\varepsilon. \] Since $\varepsilon>0$ was arbitrary, $R_n\to 0$. $\square$

Combining Lemmas 2 and 3 yields the main round-2 theorem.

Theorem. If $a_n\to\infty$ and $\tau(n)/a_n\to 0$, then \[ \sum_{n=1}^{\infty}\frac{\tau(n)}{a_1a_2\cdots a_n} \] is irrational.

A concrete nonmonotone family.

Take \[ a_{2k-1}=k^3, \qquad a_{2k}=k^2 \qquad (k\ge 1). \] Then $a_n\to\infty$, but the sequence is not monotone since $a_{2k}=k^2<a_{2k-1}=k^3$ for every $k\ge 2$. Also $\tau(n)\le n$ for all $n$, so \[ \frac{\tau(2k-1)}{a_{2k-1}}\le \frac{2k-1}{k^3}\to 0, \qquad \frac{\tau(2k)}{a_{2k}}\le \frac{2k}{k^2}\to 0. \] Thus our theorem proves that \[ \sum_{n=1}^{\infty}\frac{\tau(n)}{a_1a_2\cdots a_n} \] is irrational for this explicit nonmonotone sequence, a case not covered by Erdős--Straus.

Obstruction to any counterexample.

If a rational counterexample exists, then necessarily $R_n\not\to 0$. In particular, by contrapositive of Lemma 3, it cannot satisfy $\tau(n)/a_n\to 0$. Thus any rational counterexample must have $\tau(n)$ comparable to $a_n$ infinitely often, or at least enough mass in the scaled tails to prevent $R_n$ from tending to $0$.

6) ADVERSARIAL VERIFICATION

1. Did I use monotonicity anywhere? No. The proof only uses positivity, the integer nature of $Q_n/Q_m$ for $m\le n$, and the hypothesis $R_n\to 0$ (or the stronger concrete condition $\tau(n)/a_n\to 0$).
2. Could $R_n$ fail to be the fractional part because it is too large? Lemma 1 only asserts $\{Q_nT\}=R_n$ when $R_n<1$, and Lemma 2 uses exactly that for all sufficiently large $n$ once $R_n\to 0$.
3. Is positivity essential? Yes. The contradiction uses $0<R_n<1/r$. In the present problem all terms are positive, so this is valid.
4. Does the theorem really go beyond Round 1? Yes. The explicit family $a_{2k-1}=k^3$, $a_{2k}=k^2$ is nonmonotone, so it lies outside the known Erdős--Straus case, yet it is covered here.
5. Hidden dependence on stronger divisor estimates? None. The proof of the explicit family uses only the trivial bound $\tau(n)\le n$.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED)

The full problem remains open, but Round 2 proves a new theorem:

If $a_n\to\infty$ and $\tau(n)/a_n\to 0$, then \[ \sum_{n=1}^{\infty}\frac{\tau(n)}{a_1a_2\cdots a_n} \] is irrational.

Equivalently, any rational counterexample must violate the scaled-tail criterion; in particular it cannot have $\tau(n)=o(a_n)$. This gives a sharp obstruction and settles an explicit infinite family of nonmonotone sequences.

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 55%

9) REFERENCES

1. P. Erdős and E. G. Straus, Some number theoretic results, Pacific Journal of Mathematics 36 $1971$, 635--646.
2. T. F. Bloom (ed.), Erdős Problem #258, erdosproblems.com/258, accessed 2026-03-07.