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Math-737-A2 · revision r1 · on Math-737

gpt pro 5.2 attempt on Math-737

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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 #267
  2. 1) FORMAL RESTATEMENT
  3. 2) QUICK LITERATURE/CONTEXT CHECK (only if browsing available)
  4. 3) ATTACK PLAN
  5. 4) WORK (with full details)
  6. 4.1. A useful lemma for divisibility chains (a partial result)
  7. 4.2. Application to geometric index sets $n_k=m^k$
  8. 4.3. Why this does not settle the original problem
  9. 5) VERIFICATION / SANITY CHECK
  10. 6) FINAL
  11. 7) 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.

Lemma (Cantor-series irrationality for growing bases): Let $Q_1<Q_2<\cdots$ be positive integers with $Q_k\mid Q_{k+1}$ for all $k$, and write $q_{k+1}:=Q_{k+1}/Q_k\in\mathbb{N}$. Assume $q_{k}\to\infty$ as $k\to\infty$. Then \[ X:=\sum_{k=1}^{\infty}\frac{1}{Q_k} \] is irrational.
Proof: Set $X_N:=\sum_{k=1}^{N}\frac{1}{Q_k}$. Since $Q_k\mid Q_N$ for $k\le N$, we may write \[ X_N=\frac{P_N}{Q_N} \] for some integer $P_N$.

Let $R_N:=X-X_N=\sum_{k>N}\frac{1}{Q_k}>0$ be the tail. Multiply by $Q_N$: \[ Q_N R_N=\sum_{k>N}\frac{Q_N}{Q_k}. \] Because $Q_{N+j}=Q_N\cdot q_{N+1}q_{N+2}\cdots q_{N+j}$, we obtain \[ Q_N R_N=\sum_{j=1}^{\infty}\frac{1}{q_{N+1}q_{N+2}\cdots q_{N+j}}. \] In particular, the first term gives $Q_N R_N\le \sum_{j\ge 1} (q_{N+1})^{-j}=\frac{1}{q_{N+1}-1}$, and hence \begin{equation}\label{wsref-0-65-71-3a-51-4e-52-4e-5f-74-6f-5f-30} 0< Q_N R_N \le \frac{1}{q_{N+1}-1}\xrightarrow[N\to\infty]{} 0 \end{equation} because $q_{N+1}\to\infty$.

Now suppose for contradiction that $X\in\mathbb{Q}$, say $X=p/r$ in lowest terms. Consider the rational number \[ Q_N X - P_N = Q_N(X-X_N)=Q_N R_N. \] The left-hand side is a rational with denominator dividing $r$. Therefore either it is an integer, or else its distance to the nearest integer is at least $1/r$ (since non-integer rationals with denominator $\le r$ have fractional part a nonzero multiple of $1/r$).

For all sufficiently large $N$, the estimate (eq:QNRN_to_0) gives $0<Q_N R_N<1/r$, so $Q_N R_N$ cannot be an integer and cannot be $\ge 1/r$ away from an integer. Contradiction. Hence $X\notin\mathbb{Q}$. ∎

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.

Lemma (Divisibility of Fibonacci numbers): If $m\mid n$, then $F_m\mid F_n$.
Proof: It suffices to prove $F_m\mid F_{km}$ for all integers $k\ge 1$. We use the standard Fibonacci addition identity \begin{equation}\label{wsref-0-65-71-3a-61-64-64-69-74-69-6f-6e} F_{u+v}=F_u F_{v+1}+F_{u-1}F_v\qquad(u,v\ge 1), \end{equation} which can be proved by induction on $v$ (the $v=1$ case is $F_{u+1}=F_u+F_{u-1}$, and the induction step follows from $F_{u+(v+1)}=F_{u+v}+F_{u+v-1}$ and applying (eq:addition) to $v$ and $v-1$).

Fix $m$. For $k=1$, $F_m\mid F_{m}$ holds. Assume $F_m\mid F_{km}$. Apply (eq:addition) with $u=km$ and $v=m$: \[ F_{(k+1)m}=F_{km+m}=F_{km}F_{m+1}+F_{km-1}F_m. \] Both terms on the right are divisible by $F_m$ (the first by the induction hypothesis, the second trivially), hence $F_m\mid F_{(k+1)m}$. This completes the induction. ∎

Proposition (A partial answer: $n_k=m^k$ gives an irrational sum): Fix an integer $m\ge 2$ and set $n_k:=m^k$. Then \[ \sum_{k=1}^{\infty}\frac{1}{F_{m^k}} \] is irrational.
Proof: Set $Q_k:=F_{m^k}$. Since $m^k\mid m^{k+1}$, Lemma Lemma (Divisibility of Fibonacci numbers) gives $Q_k\mid Q_{k+1}$.

Let $q_{k+1}:=Q_{k+1}/Q_k$. Using the standard asymptotic $F_n\sim \varphi^n/\sqrt{5}$ (with $\varphi=(1+\sqrt5)/2$), we have \[ q_{k+1}=\frac{F_{m^{k+1}}}{F_{m^k}}\asymp \varphi^{m^{k+1}-m^k}=\varphi^{(m-1)m^k}\xrightarrow[k\to\infty]{}\infty. \] Therefore the hypotheses of Lemma Lemma (Cantor-series irrationality for growing bases) hold, and the sum $\sum_{k\ge 1} 1/Q_k=\sum_{k\ge 1}1/F_{m^k}$ is irrational. ∎

Remark: The special case $m=2$ recovers the irrationality of $\sum_k 1/F_{2^k}$ (which was originally established by Good and by Bicknell--Hoggatt). The argument above shows irrationality more generally for any geometric subsequence $m^k$.

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.
The partial result above indicates that divisibility chains force irrationality, but extending to general lacunary sequences remains a substantial gap.