Erdős Problem #267

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.

Erdős Problem #267: independent continuation

1) OBJECTIVE AND SOURCE AUDIT

2) WORK: A SHARPER PRODUCT CRITERION

The exact constant satisfies $\kappa>1$, so this is a modest strengthening of the supplied sufficient criterion. It is meaningful for index sequences whose first omitted index is only linearly below the previous-index sum. For indices asymptotic to $c^N$ with fixed $1<c<2$, however, \[ n_{N+1}-\sum_{j\le N}n_j \sim n_{N+1}\frac{c-2}{c-1}, \] which is negative on an exponential scale; no linear correction in $N$ can repair the product-denominator method in that regime.

3) VERIFICATION AND FINAL

8) COMPLETION ESTIMATE

Erdős Problem #267

1) FORMAL RESTATEMENT

Let $(F_n)_{n\ge 1}$ be the Fibonacci numbers defined by $F_1=F_2=1$ and $F_{n+1}=F_n+F_{n-1}$ for $n\ge 2$.

Given an increasing sequence of indices $1\le n_1<n_2<\cdots$ such that \[ \frac{n_{k+1}}{n_k}\ge c>1\quad\text{for all }k, \] must the subseries \[ S:=\sum_{k=1}^{\infty}\frac{1}{F_{n_k}} \] be irrational?

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

• The special lacunary case $n_k=2^k$ was proved to give an irrational sum by Good $1974$ and by Bicknell--Hoggatt $1976$.
• The full reciprocal Fibonacci constant $\sum_{n\ge 1} 1/F_n$ was proved irrational by André--Jeannin $1989$.
• The general "ratio bounded away from 1" condition $n_{k+1}/n_k\ge c>1$ is significantly weaker than divisibility assumptions (e.g. $n_k\mid n_{k+1}$), and the problem is listed as open in modern compilations.

3) ATTACK PLAN

1. Isolate a tractable strengthening of the hypothesis (e.g. $n_k\mid n_{k+1}$) under which the sum can be rewritten as a Cantor-type series.
2. Prove irrationality in that stronger setting using a general "mixed radix" approximation lemma.
3. Record that the original problem removes divisibility, and explain why the Cantor-series proof no longer applies directly.

4) WORK (with full details)

4.1. A useful lemma for divisibility chains (a partial result)

We begin with a general irrationality lemma.

4.2. Application to geometric index sets $n_k=m^k$

We now show that many lacunary Fibonacci subseries are irrational under a stronger hypothesis than in the problem.

4.3. Why this does not settle the original problem

Lemma Lemma (Cantor-series irrationality for growing bases) crucially used a divisibility chain $Q_k\mid Q_{k+1}$, which is guaranteed if $n_k\mid n_{k+1}$ (by Lemma Lemma (Divisibility of Fibonacci numbers)). The Erdős hypothesis $n_{k+1}/n_k\ge c>1$ does not imply any such divisibility, so the denominators $F_{n_k}$ need not nest.

Without nesting, the partial sums $\sum_{k\le N} 1/F_{n_k}$ do not share a single "natural" denominator $Q_N$ dividing all others, so the clean modular obstruction in Lemma Lemma (Cantor-series irrationality for growing bases) is unavailable.

5) VERIFICATION / SANITY CHECK

• The bound $Q_N R_N\le 1/(q_{N+1}-1)$ is obtained by comparing the tail to a geometric series using $q_{N+j}\ge q_{N+1}$ for $j\ge 1$ (valid since $q_k\ge 2$ and $q_k\to\infty$ so eventually nondecreasing; if not monotone, one can instead use $Q_N R_N\le \sum_{j\ge 1} 1/q_{N+1}^j$ after replacing $q_{N+1}$ by $\min_{j\ge 1} q_{N+j}$).
• In Proposition prop:mk, the fact $q_{k+1}\to\infty$ follows from exponential growth of Fibonacci numbers.
• The argument for divisibility $F_m\mid F_{km}$ is elementary and uses only the addition identity.

6) FINAL

UNRESOLVED.

• (i) What is proved here: under the stronger hypothesis $n_k=m^k$ (or more generally $n_k\mid n_{k+1}$ with rapidly growing quotients), the sum $\sum 1/F_{n_k}$ is irrational.
• (ii) What remains open: the original statement for arbitrary sequences with only the ratio gap condition $n_{k+1}/n_k\ge c>1$ (no divisibility).
• (iii) Most plausible next step: develop an approximation/irrationality criterion that does not rely on nested denominators, perhaps exploiting gcd-structure $\gcd(F_m,F_n)=F_{\gcd(m,n)}$ or using lower bounds on least common multiples of $F_{n_k}$.
• (iv) Completion estimate: proving irrationality in the full generality likely needs new number-theoretic input beyond mixed-radix arguments; alternatively, a counterexample would require constructing a sparse index set with an unexpectedly rational reciprocal sum.

7) COMPLETION ESTIMATE

To completely resolve Erdős #267 one would need either:

1. a general irrationality criterion covering all lacunary Fibonacci subseries with $n_{k+1}/n_k\ge c>1$, or
2. an explicit construction of such a lacunary sequence $(n_k)$ yielding a rational sum.