gpt pro 5.4 attempt on Math-749

Erdős Problem #295

1) ROUND-2 OBJECTIVE

I pursue path (C): correct a concrete Round-1 error and isolate a sharp obstruction to the Round-1 proof strategy.

The Round-1 proof correctly established the necessary condition \[ 1\le \sum_{i=0}^{k-1}\frac1{N+i}, \] but the final algebraic simplification contained an off-by-one error. After correcting that, I sharpen the analysis of the same necessary condition and prove that this entire harmonic-comparison strategy can at best yield a bounded additive correction to $(e-1)N$. That does not settle the problem, but it cleanly identifies why the Round-1 approach cannot prove \[ k(N)-(e-1)N\to\infty. \]

2) Round-1 FOUNDATION USED

From Round 1 I use only the following valid ingredient.

Round-1 Proposition. If \[ N\le n_1<n_2<\cdots<n_k \qquad\mbox{and}\qquad \frac1{n_1}+\cdots+\frac1{n_k}=1, \] then \[ 1\le \sum_{i=0}^{k-1}\frac1{N+i}. \]

I do not use the Round-1 corollary \[ k(N)\ge (e-1)(N-1)+1, \] because it came from a faulty algebraic simplification; I correct that below.

3) NEW INSIGHT / TOOL (ROUND-2)

There are two new ingredients.

• First, a correction: the same integral argument from Round 1 gives only \[ k(N)\ge (e-1)N-(e-1), \] not $(e-1)(N-1)+1$.
• Second, define the harmonic threshold \[ L(N):=\min\left\{k\ge 1:\ \sum_{i=0}^{k-1}\frac1{N+i}\ge 1\right\}. \] Then every valid representation must satisfy $k(N)\ge L(N)$, and \[ L(N)=(e-1)N-\frac{e-1}{2}+O(1). \] Hence any argument that uses only the Round-1 necessary condition can never prove divergence of $k(N)-(e-1)N$.

4) ATTACK PLAN (ROUND-2)

The precise Round-1 gap is this:

• the Round-1 necessary condition was correct;
• the final lower bound had an algebra slip;
• even after correcting that slip, the method is too weak to imply any unbounded additive term.

So I will:

• correct the Round-1 lower bound;
• identify the exact threshold $L(N)$ coming from the harmonic method;
• show asymptotically that $L(N)$ differs from $(e-1)N$ by only a bounded term;
• extend the exact computation of $k(N)$ from $N\le 3$ to $N\le 11$.

This addresses the Round-1 obstacle directly: it proves that the obstacle is structural, not just a matter of doing the same calculation more carefully.

5) WORK (ROUND-2)

Step 1: correcting the Round-1 algebra.

By the Round-1 proposition, \[ 1\le \sum_{i=0}^{k-1}\frac1{N+i}. \] Since $x\mapsto 1/x$ decreases, \[ \frac1{N+i}\le \int_{N+i-1}^{N+i}\frac{dx}{x}, \] and therefore \[ 1\le \sum_{i=0}^{k-1}\frac1{N+i} \le \int_{N-1}^{N+k-1}\frac{dx}{x} = \log\!\left(\frac{N+k-1}{N-1}\right). \] Thus \[ \frac{N+k-1}{N-1}\ge e, \] so \[ N+k-1\ge e(N-1), \] and hence \[ k\ge e(N-1)-N+1=(e-1)N-(e-1). \]

This is the correct conclusion from the Round-1 integral comparison.

Step 2: the exact harmonic threshold.

Define \[ H_m:=1+\frac12+\cdots+\frac1m, \] and \[ L(N):=\min\{k\ge 1:\ H_{N+k-1}-H_{N-1}\ge 1\}. \] Then the Round-1 proposition is equivalent to

Lemma 1. For every $N\ge 1$, \[ k(N)\ge L(N). \]

Proof. Immediate from \[ 1\le \sum_{i=0}^{k-1}\frac1{N+i}=H_{N+k-1}-H_{N-1}. \]

Step 3: asymptotics of $L(N)$.

I use the standard harmonic-number expansion \[ H_n=\log n+\gamma+\frac1{2n}+O\!\left(\frac1{n^2}\right) \qquad (n\to\infty). \] Equivalently, \[ H_n=\log\!\left(n+\frac12\right)+\gamma+O\!\left(\frac1{n^2}\right), \] because \[ \log\!\left(n+\frac12\right)=\log n+\frac1{2n}+O\!\left(\frac1{n^2}\right). \]

Therefore, uniformly for $k=O(N)$, \[ H_{N+k-1}-H_{N-1} = \log\!\left(\frac{N+k-\frac12}{\,N-\frac12\,}\right) + O\!\left(\frac1{N^2}\right). \]

Set \[ r_N:=(e-1)\left(N-\frac12\right)=(e-1)N-\frac{e-1}{2}. \] Then for any fixed real $C$, \[ N+(r_N+C)-\frac12=e\left(N-\frac12\right)+C, \] so \[ \frac{N+(r_N+C)-\frac12}{N-\frac12} = e+\frac{C}{N-\frac12}. \] Hence \[ \log\!\left(\frac{N+(r_N+C)-\frac12}{N-\frac12}\right) = 1+\frac{C}{e(N-\frac12)}+O\!\left(\frac1{N^2}\right). \]

Because the crude bound above already shows $L(N)=O(N)$, the threshold regime indeed has $k=O(N)$. Choosing $C=2$ and $C=-2$, the last display shows that for all sufficiently large $N$, \[ H_{N+\lfloor r_N-2\rfloor-1}-H_{N-1}<1, \qquad H_{N+\lceil r_N+2\rceil-1}-H_{N-1}>1. \] By the definition of $L(N)$ this implies \[ r_N-2 < L(N) \le r_N+3 \] for all sufficiently large $N$. Therefore \[ L(N)=r_N+O(1)=(e-1)N-\frac{e-1}{2}+O(1). \]

Thus we have proved:

Theorem 2. As $N\to\infty$, \[ L(N)=(e-1)N-\frac{e-1}{2}+O(1). \]

Corollary 3 (sharp obstruction to the Round-1 method). Any lower bound obtained solely from the necessary condition \[ 1\le \sum_{i=0}^{k-1}\frac1{N+i} \] can improve $(e-1)N$ by at most a bounded additive term. In particular, this method cannot prove \[ k(N)-(e-1)N\to\infty. \]

Proof. Such an argument can give at best $k(N)\ge L(N)$, and Theorem 2 shows \[ L(N)-(e-1)N=-\frac{e-1}{2}+O(1), \] which is bounded.

Step 4: exact computation for $N\le 11$.

I extended the Round-1 computation from $N\le 3$ to $N\le 11$.

The search is rigorous and finite. If one is trying to represent a positive rational remainder $\rho$ as a sum of $r$ distinct reciprocals with next denominator $d$, then necessarily \[ d\ge \left\lceil \frac1{\rho}\right\rceil \qquad\mbox{and}\qquad d\le \left\lfloor \frac{r}{\rho}\right\rfloor, \] since even $r$ copies of $1/d$ cannot reach $\rho$ when $d>r/\rho$. Moreover, if \[ \frac1d+\frac1{d+1}+\cdots+\frac1{d+r-1}<\rho, \] then the branch is impossible. This gives an exhaustive finite recursion.

The exact values obtained are: \[ \begin{array}{c|ccccccccccc} N & 1&2&3&4&5&6&7&8&9&10&11\\ \hline L(N) & 1&3&5&7&8&10&12&13&15&17&19\\ k(N) & 1&3&5&8&10&11&13&15&17&19&21 \end{array} \]

For example, the search found \[ 1=\sum_{n=11}^{28}\frac1n+\frac1{561}+\frac1{71820}+\frac1{1315600}, \] so $k(11)\le 21$, while exhaustive search shows that no representation exists with $19$ or $20$ terms and denominators all at least $11$. Hence $k(11)=21$.

Similarly, \[ 1=\sum_{n=10}^{22}\frac1n+\frac1{24}+\frac1{26}+\frac1{28}+\frac1{45} +\frac1{11088}+\frac1{9699690}, \] so $k(10)=19$.

6) ADVERSARIAL VERIFICATION

• Round-1 error check. The corrected inequality \[ k\ge e(N-1)-N+1=(e-1)N-(e-1) \] comes from a direct rearrangement of $N+k-1\ge e(N-1)$. The discarded Round-1 constant was indeed an algebra mistake.
• Is the obstruction too strong? No. I am not claiming that the original Erdős problem is false; I am only proving that the particular "compare with the first $k$ reciprocals" strategy cannot prove divergence.
• Uniformity of the asymptotic expansion. The formula for $H_{N+k-1}-H_{N-1}$ is used only in the range $k=O(N)$, which is already guaranteed by the crude integral lower bound together with the trivial upper comparison defining $L(N)$.
• Computational exactness. The recursion is finite because the next denominator is bounded above by $\lfloor r/\rho\rfloor$ at every stage. Thus negative outcomes are genuine exhaustive nonexistence results, not heuristic failures to find a solution.

7) FINAL (EXACTLY ONE)

UNRESOLVED (BUT STRICTLY ADVANCED).

Round 2 does not settle whether \[ k(N)-(e-1)N\to\infty, \] but it makes two strict advances:

• it corrects the Round-1 lower bound;
• it proves that the entire Round-1 harmonic-comparison strategy is inherently incapable of proving the desired divergence.

In addition, it extends the exact computation of $k(N)$ from $N\le 3$ to $N\le 11$.

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 40%

9) REFERENCES

1. T. F. Bloom, Erdős Problem #295, erdosproblems.com/295, accessed 2026-03-07.
2. The estimate \[ -c<k(N)-(e-1)N\ll \frac{N}{\log N} \] was checked from the problem page above.
3. The harmonic-number asymptotic used above is the standard Euler--Maclaurin expansion.