gpt pro 5.4 attempt on Math-737

Erdős Problem #267 (Round 2)

1. ROUND-2 OBJECTIVE

I pursue (C) obstruction/correction. Round 1 treated only divisibility chains such as $n_k=m^k$. In this round I correct an omission: the case $n_{k+1}/n_k\ge 2$ is already known in the literature, and in fact it has a short self-contained proof that does not require divisibility. I will prove a slightly more general sufficient condition and deduce the whole $c\ge 2$ range.

2. Round-1 FOUNDATION USED

I use only the most basic Round 1 ingredients:

1. the notation $S=\sum_k 1/F_{n_k}$;
2. the idea of comparing a rational approximation denominator against the tail of the series;
3. the standard Binet-growth heuristic for Fibonacci numbers.

I will not reuse the Round 1 divisibility-chain lemma; the new proof avoids divisibility entirely.

3. NEW INSIGHT / TOOL (ROUND-2)

The new observation is that one can multiply the $N$th partial sum by the full product \[ Q_N:=\prod_{j=1}^{N}F_{n_j} \] instead of by a nested denominator. If the indices are sufficiently lacunary, then the tail \[ R_N:=\sum_{k>N}\frac1{F_{n_k}} \] is so small that $Q_N R_N\to 0$. This yields irrationality immediately.

This gives the following general criterion.

Criterion. If \[ n_{N+1}-\sum_{j=1}^{N} n_j + N \to \infty, \] then $\sum_k 1/F_{n_k}$ is irrational.

The Erdős hypothesis with $c\ge 2$ is an immediate corollary.

4. ATTACK PLAN (ROUND-2)

The exact gap after Round 1 was the lack of any result beyond divisibility chains. I now prove:

Claim A. If $n_{N+1}-\sum_{j\le N}n_j+N\to\infty$, then $\sum_k 1/F_{n_k}$ is irrational.

Claim B. If $n_{k+1}/n_k\ge c\ge 2$ for all $k$, then the hypothesis of Claim A holds; hence the whole range $c\ge 2$ is settled.

This overcomes the main Round 1 obstacle because it eliminates the need for denominator divisibility.

5. WORK (ROUND-2)

Let \[ S:=\sum_{k=1}^{\infty}\frac{1}{F_{n_k}}, \qquad S_N:=\sum_{k=1}^{N}\frac{1}{F_{n_k}}, \qquad R_N:=S-S_N. \] Also define \[ Q_N:=\prod_{j=1}^{N}F_{n_j}. \] Then $Q_N S_N$ is an integer.

Lemma 5.1. For every integer $m\ge 1$, \[ \varphi^{m-2}\le F_m\le \varphi^{m-1}, \] where $\varphi=(1+\sqrt5)/2$.

Proof. We argue by induction on $m$. The bounds are true for $m=1,2$ because \[ \varphi^{-1}\le 1\le 1, \qquad 1\le 1\le \varphi. \] If they hold for $m$ and $m-1$, then using $F_{m+1}=F_m+F_{m-1}$ and $\varphi^2=\varphi+1$, \[ F_{m+1}\le \varphi^{m-1}+\varphi^{m-2}=\varphi^m, \] and similarly \[ F_{m+1}\ge \varphi^{m-2}+\varphi^{m-3}=\varphi^{m-1}. \] This closes the induction. $\square$

Theorem 5.2. Assume that \[ \Delta_N:=n_{N+1}-\sum_{j=1}^{N}n_j+N \to \infty. \] Then the series \[ \sum_{k=1}^{\infty}\frac{1}{F_{n_k}} \] is irrational.

Proof. Lemma 5.1 gives \[ Q_N=\prod_{j=1}^{N}F_{n_j} \le \prod_{j=1}^{N}\varphi^{n_j-1} =\varphi^{\sum_{j=1}^{N}n_j-N}. \] It also gives \[ \frac{1}{F_{n_k}}\le \varphi^{2-n_k} \qquad\mbox{for all }k. \] Hence \[ R_N\le \sum_{k>N}\varphi^{2-n_k}. \] Since $(n_k)$ is strictly increasing, we have $n_{N+m}\ge n_{N+1}+m-1$ for $m\ge 1$, so \[ R_N \le \sum_{m=1}^{\infty}\varphi^{2-(n_{N+1}+m-1)} =\varphi^{3-n_{N+1}}\sum_{m=1}^{\infty}\varphi^{-m} \le C\varphi^{-n_{N+1}} \] for an absolute constant $C>0$. Therefore \[ Q_N R_N \le C\varphi^{\sum_{j=1}^{N}n_j-N-n_{N+1}} = C\varphi^{-\Delta_N} \to 0. \] Now suppose, for contradiction, that $S=p/q$ is rational. Then \[ qQ_N R_N=qQ_N S-qQ_N S_N. \] The right-hand side is a positive integer, because $Q_N S_N\in\mathbf Z$ and $R_N>0$. But the left-hand side tends to $0$, impossible. Therefore $S$ is irrational. $\square$

Corollary 5.3. If \[ \frac{n_{k+1}}{n_k}\ge 2 \qquad\mbox{for all }k, \] then \[ \sum_{k=1}^{\infty}\frac{1}{F_{n_k}} \] is irrational.

Proof. From $n_{k+1}\ge 2n_k$ we get, for each fixed $N$, \[ n_j\le \frac{n_{N+1}}{2^{N+1-j}} \qquad (1\le j\le N). \] Therefore \[ \sum_{j=1}^{N}n_j \le n_{N+1}\sum_{m=1}^{N}2^{-m} = n_{N+1}(1-2^{-N}). \] Hence \[ \Delta_N = n_{N+1}-\sum_{j=1}^{N}n_j+N \ge 2^{-N}n_{N+1}+N \ge n_1+N \to\infty. \] Now apply Theorem 5.2. $\square$

Corollary 5.4. The Erdős question in #267 has an affirmative answer throughout the full range $c\ge 2$.

This recovers the known $c\ge 2$ theorem and shows exactly where the proof breaks when $1<c<2$: the product denominator $Q_N$ becomes too large compared with the first omitted Fibonacci denominator.

6. ADVERSARIAL VERIFICATION

1. No divisibility hidden anywhere. The proof uses only the product $Q_N$, never a least common multiple or a divisibility chain.
2. Boundary case $c=2$. The estimate for $\Delta_N$ still tends to infinity because the extra $+N$ term remains, and in fact $\Delta_N\ge n_1+N$.
3. Rationality contradiction. The expression $qQ_N R_N$ is a positive integer, not merely a rational bounded away from $0$, because $Q_N S_N$ is an integer by construction.
4. What remains open. For $1<c<2$, the quantity $\Delta_N$ need not tend to infinity, so this particular product-denominator strategy genuinely stops there.

7. FINAL

UNRESOLVED (BUT STRICTLY ADVANCED)

Round 2 corrects the Round 1 picture in two ways.

1. It identifies the already-settled range: the problem is known for $c\ge 2$.
2. It gives a short self-contained proof of the stronger criterion \[ n_{N+1}-\sum_{j=1}^{N}n_j+N\to\infty \implies \sum_k \frac1{F_{n_k}}\ \mbox{irrational}, \] of which the whole range $c\ge 2$ is an immediate corollary.

The original Erdős problem is therefore reduced to the genuinely difficult regime \[ 1<c<2. \]

8. COMPLETION ESTIMATE

COMPLETION: 78%

9. REFERENCES

1. C. Badea, A theorem on irrationality of infinite series and applications, Acta Arith. 63 $1993$, 313--323.