Erdős Problem #260

Erdős Problem #260

1) ROUND-2 OBJECTIVE

I pursue path (C) in the fail-safe sense: not a full proof of the original problem, but a strictly stronger theorem than Round 1 together with a sharper identification of the remaining obstruction. Round 1 proved irrationality under the condition \[ a_{n+1}-a_n-\lfloor \log_2 a_{n+1}\rfloor \longrightarrow \infty. \] The new goal is to weaken this to a mere limsup condition.

2) Round-1 FOUNDATION USED

I use only the Round 1 binary-support viewpoint: \[ \frac{a_n}{2^{a_n}} \] contributes binary digits only in the interval \[ [a_n-\lfloor \log_2 a_n\rfloor,\; a_n]. \] Round 1 used eventual disjointness of these intervals to get a no-carry argument.

3) NEW INSIGHT / TOOL (ROUND-2)

The new tool is a tail-stability lemma for binary digits: if a dyadic partial sum has denominator $2^m$ and the remaining tail is less than $2^{-M}$, then the first $M-1$ binary digits are frozen, regardless of later carries.

This bypasses the Round 1 no-carry requirement entirely. I only need infinitely many large gaps, not eventual gap separation.

4) ATTACK PLAN (ROUND-2)

The exact Round 1 gap was this: eventual disjointness of digit blocks was used to control carries. I replace that by a quantitative tail estimate.

Define \[ L_n:=\lfloor \log_2 a_n\rfloor,\qquad \Delta_n:=a_{n+1}-L_{n+1}-a_n-1. \] Thus $\Delta_n$ is the number of binary positions strictly between the last possible nonzero digit coming from the $n$th summand and the first possible nonzero digit coming from the $(n+1)$st summand.

The claim to prove is:

New theorem. If \[ \limsup_{n\to\infty}\Delta_n=\infty, \] then \[ \sum_{n=1}^{\infty}\frac{a_n}{2^{a_n}} \] is irrational.

This strictly strengthens the Round 1 theorem, which required $\Delta_n\to\infty$.

5) WORK (ROUND-2)

Let \[ S:=\sum_{n=1}^{\infty}\frac{a_n}{2^{a_n}},\qquad P_n:=\sum_{k=1}^{n}\frac{a_k}{2^{a_k}},\qquad T_n:=S-P_n=\sum_{k>n}\frac{a_k}{2^{a_k}}. \]

Step 1: a sharp upper bound for the tail.

Since $a_{n+1}<a_{n+2}<\cdots$, we have \[ T_n\leq \sum_{m=a_{n+1}}^{\infty}\frac{m}{2^m}. \] Using the standard identity \[ \sum_{m=N}^{\infty}\frac{m}{2^m}=\frac{N+1}{2^{N-1}} \qquad (N\geq 1), \] or equivalently \[ \sum_{m=N+1}^{\infty}\frac{m}{2^m}=\frac{N+2}{2^N}, \] we get \[ T_n\leq \frac{a_{n+1}+2}{2^{a_{n+1}}}. \] Now $a_{n+1}<2^{L_{n+1}+1}$, hence $a_{n+1}+2<2^{L_{n+1}+2}$. Therefore \[ T_n<2^{L_{n+1}+2-a_{n+1}}. \] Because \[ a_{n+1}=a_n+1+\Delta_n+L_{n+1}, \] this becomes \[ T_n<2^{-(a_n+\Delta_n-1)}. \tag{260.1} \]

Step 2: long zero runs in the binary expansion of $S$.

The partial sum $P_n$ is a dyadic rational with denominator dividing $2^{a_n}$. Hence in its terminating binary expansion, all digits in positions strictly after $a_n$ are $0$.

From $260.1$, adding $T_n$ can change only binary digits in positions at least $a_n+\Delta_n-1$. Therefore the binary digits of $S$ in the positions \[ a_n+1,\; a_n+2,\; \dots,\; a_n+\Delta_n-2 \] agree with those of $P_n$, hence are all $0$.

So whenever $\Delta_n\geq 2$, the binary expansion of $S$ contains a zero run of length at least \[ \Delta_n-2. \] Since $\limsup \Delta_n=\infty$, the binary expansion of $S$ contains arbitrarily long zero runs.

Step 3: a rational number with arbitrarily long zero runs must be dyadic.

Take the standard binary expansion, i.e. the one not ending in an infinite tail of $1$'s. If a rational number is not dyadic, then its standard binary expansion is eventually periodic but not eventually zero. If the eventual period has length $q$, then every sufficiently late block of $q$ consecutive digits contains at least one $1$; otherwise one whole period would be all zero, forcing termination. Hence the lengths of zero runs are bounded.

Therefore a rational number whose standard binary expansion has arbitrarily long zero runs must be dyadic.

So, to finish, it remains only to rule out the possibility that $S$ is dyadic.

Step 4: $S$ is not dyadic.

Assume for contradiction that \[ S=\frac{u}{2^M} \] for some integers $u,M\geq 0$.

Choose $n$ so large that $a_n\geq M$ and $\Delta_n\geq 2$; this is possible because $a_n\to\infty$ and $\limsup \Delta_n=\infty$.

Then $P_n$ has denominator dividing $2^{a_n}$, and so does $S$. Hence \[ T_n=S-P_n \] is a nonnegative multiple of $2^{-a_n}$.

But $T_n>0$ because there are infinitely many positive summands after the $n$th one, while by $260.1$ and $\Delta_n\geq 2$ we have \[ 0<T_n<2^{-a_n}. \] This is impossible, because the only multiple of $2^{-a_n}$ lying in the interval $(0,2^{-a_n})$ is none at all.

This contradiction shows that $S$ is not dyadic. Combined with Step 3, it follows that $S$ is irrational.

Conclusion. If \[ \limsup_{n\to\infty}\left(a_{n+1}-a_n-\lfloor \log_2 a_{n+1}\rfloor-1\right)=\infty, \] then \[ \sum_{n=1}^{\infty}\frac{a_n}{2^{a_n}} \] is irrational.

6) ADVERSARIAL VERIFICATION

• Carry issue. The proof never assumes no-carry. The only input is that the tail is numerically too small to alter early binary digits.
• Terminating-binary loophole. Arbitrarily long zero runs alone do not exclude dyadic rationals; this is why Step 4 is separate.
• Quantifiers. I only need infinitely many indices with large $\Delta_n$, not eventual control. This is exactly why the hypothesis is a $\limsup$ condition.
• Strict advance over Round 1. Round 1 required $\Delta_n\to\infty$; the new theorem needs only $\limsup \Delta_n=\infty$.

7) FINAL (EXACTLY ONE)

UNRESOLVED (BUT STRICTLY ADVANCED)

Round 2 proves a strictly stronger irrationality criterion than Round 1: \[ \limsup_{n\to\infty}\left(a_{n+1}-a_n-\lfloor \log_2 a_{n+1}\rfloor-1\right)=\infty \] already suffices. The genuine remaining obstruction is therefore concentrated in the regime where \[ a_{n+1}-a_n-\lfloor \log_2 a_{n+1}\rfloor \] stays bounded above, so the tail can repeatedly interact with the earlier binary digits.

8) COMPLETION ESTIMATE

COMPLETION: 50%

9) REFERENCES

• [EP260] T. F. Bloom, Erdős Problem #260, problem page, accessed 2026-03-07.
• [Er88c] P. Erdős, On the irrationality of certain series: problems and results, 1988; see p. 103 for Problem #260 and the discussion of stronger sufficient hypotheses.

Erdős Problem #260

1) ROUND-2 OBJECTIVE

I pursue path (C) in the fail-safe sense: not a full proof of the original problem, but a strictly stronger theorem than Round 1 together with a sharper identification of the remaining obstruction. Round 1 proved irrationality under the condition \[ a_{n+1}-a_n-\lfloor \log_2 a_{n+1}\rfloor \longrightarrow \infty. \] The new goal is to weaken this to a mere limsup condition.

2) Round-1 FOUNDATION USED

I use only the Round 1 binary-support viewpoint: \[ \frac{a_n}{2^{a_n}} \] contributes binary digits only in the interval \[ [a_n-\lfloor \log_2 a_n\rfloor,\; a_n]. \] Round 1 used eventual disjointness of these intervals to get a no-carry argument.

3) NEW INSIGHT / TOOL (ROUND-2)

The new tool is a tail-stability lemma for binary digits: if a dyadic partial sum has denominator $2^m$ and the remaining tail is less than $2^{-M}$, then the first $M-1$ binary digits are frozen, regardless of later carries.

This bypasses the Round 1 no-carry requirement entirely. I only need infinitely many large gaps, not eventual gap separation.

4) ATTACK PLAN (ROUND-2)

The exact Round 1 gap was this: eventual disjointness of digit blocks was used to control carries. I replace that by a quantitative tail estimate.

Define \[ L_n:=\lfloor \log_2 a_n\rfloor,\qquad \Delta_n:=a_{n+1}-L_{n+1}-a_n-1. \] Thus $\Delta_n$ is the number of binary positions strictly between the last possible nonzero digit coming from the $n$th summand and the first possible nonzero digit coming from the $(n+1)$st summand.

The claim to prove is:

New theorem. If \[ \limsup_{n\to\infty}\Delta_n=\infty, \] then \[ \sum_{n=1}^{\infty}\frac{a_n}{2^{a_n}} \] is irrational.

This strictly strengthens the Round 1 theorem, which required $\Delta_n\to\infty$.

5) WORK (ROUND-2)

Let \[ S:=\sum_{n=1}^{\infty}\frac{a_n}{2^{a_n}},\qquad P_n:=\sum_{k=1}^{n}\frac{a_k}{2^{a_k}},\qquad T_n:=S-P_n=\sum_{k>n}\frac{a_k}{2^{a_k}}. \]

Step 1: a sharp upper bound for the tail.

Since $a_{n+1}<a_{n+2}<\cdots$, we have \[ T_n\leq \sum_{m=a_{n+1}}^{\infty}\frac{m}{2^m}. \] Using the standard identity \[ \sum_{m=N}^{\infty}\frac{m}{2^m}=\frac{N+1}{2^{N-1}} \qquad (N\geq 1), \] or equivalently \[ \sum_{m=N+1}^{\infty}\frac{m}{2^m}=\frac{N+2}{2^N}, \] we get \[ T_n\leq \frac{a_{n+1}+2}{2^{a_{n+1}}}. \] Now $a_{n+1}<2^{L_{n+1}+1}$, hence $a_{n+1}+2<2^{L_{n+1}+2}$. Therefore \[ T_n<2^{L_{n+1}+2-a_{n+1}}. \] Because \[ a_{n+1}=a_n+1+\Delta_n+L_{n+1}, \] this becomes \[ T_n<2^{-(a_n+\Delta_n-1)}. \tag{260.1} \]

Step 2: long zero runs in the binary expansion of $S$.

The partial sum $P_n$ is a dyadic rational with denominator dividing $2^{a_n}$. Hence in its terminating binary expansion, all digits in positions strictly after $a_n$ are $0$.

From $260.1$, adding $T_n$ can change only binary digits in positions at least $a_n+\Delta_n-1$. Therefore the binary digits of $S$ in the positions \[ a_n+1,\; a_n+2,\; \dots,\; a_n+\Delta_n-2 \] agree with those of $P_n$, hence are all $0$.

So whenever $\Delta_n\geq 2$, the binary expansion of $S$ contains a zero run of length at least \[ \Delta_n-2. \] Since $\limsup \Delta_n=\infty$, the binary expansion of $S$ contains arbitrarily long zero runs.

Step 3: a rational number with arbitrarily long zero runs must be dyadic.

Take the standard binary expansion, i.e. the one not ending in an infinite tail of $1$'s. If a rational number is not dyadic, then its standard binary expansion is eventually periodic but not eventually zero. If the eventual period has length $q$, then every sufficiently late block of $q$ consecutive digits contains at least one $1$; otherwise one whole period would be all zero, forcing termination. Hence the lengths of zero runs are bounded.

Therefore a rational number whose standard binary expansion has arbitrarily long zero runs must be dyadic.

So, to finish, it remains only to rule out the possibility that $S$ is dyadic.

Step 4: $S$ is not dyadic.

Assume for contradiction that \[ S=\frac{u}{2^M} \] for some integers $u,M\geq 0$.

Choose $n$ so large that $a_n\geq M$ and $\Delta_n\geq 2$; this is possible because $a_n\to\infty$ and $\limsup \Delta_n=\infty$.

Then $P_n$ has denominator dividing $2^{a_n}$, and so does $S$. Hence \[ T_n=S-P_n \] is a nonnegative multiple of $2^{-a_n}$.

But $T_n>0$ because there are infinitely many positive summands after the $n$th one, while by $260.1$ and $\Delta_n\geq 2$ we have \[ 0<T_n<2^{-a_n}. \] This is impossible, because the only multiple of $2^{-a_n}$ lying in the interval $(0,2^{-a_n})$ is none at all.

This contradiction shows that $S$ is not dyadic. Combined with Step 3, it follows that $S$ is irrational.

Conclusion. If \[ \limsup_{n\to\infty}\left(a_{n+1}-a_n-\lfloor \log_2 a_{n+1}\rfloor-1\right)=\infty, \] then \[ \sum_{n=1}^{\infty}\frac{a_n}{2^{a_n}} \] is irrational.

6) ADVERSARIAL VERIFICATION

• Carry issue. The proof never assumes no-carry. The only input is that the tail is numerically too small to alter early binary digits.
• Terminating-binary loophole. Arbitrarily long zero runs alone do not exclude dyadic rationals; this is why Step 4 is separate.
• Quantifiers. I only need infinitely many indices with large $\Delta_n$, not eventual control. This is exactly why the hypothesis is a $\limsup$ condition.
• Strict advance over Round 1. Round 1 required $\Delta_n\to\infty$; the new theorem needs only $\limsup \Delta_n=\infty$.

7) FINAL (EXACTLY ONE)

UNRESOLVED (BUT STRICTLY ADVANCED)

Round 2 proves a strictly stronger irrationality criterion than Round 1: \[ \limsup_{n\to\infty}\left(a_{n+1}-a_n-\lfloor \log_2 a_{n+1}\rfloor-1\right)=\infty \] already suffices. The genuine remaining obstruction is therefore concentrated in the regime where \[ a_{n+1}-a_n-\lfloor \log_2 a_{n+1}\rfloor \] stays bounded above, so the tail can repeatedly interact with the earlier binary digits.

8) COMPLETION ESTIMATE

COMPLETION: 50%

9) REFERENCES

• [EP260] T. F. Bloom, Erdős Problem #260, problem page, accessed 2026-03-07.
• [Er88c] P. Erdős, On the irrationality of certain series: problems and results, 1988; see p. 103 for Problem #260 and the discussion of stronger sufficient hypotheses.

OBJECTIVE AND REVIEW OF ALL VERSIONS

WORK: CORRECTED TAIL AND RATIONALITY OBSTRUCTION

ADVERSARIAL VERIFICATION AND REMAINING GAP

SOURCES CHECKED

FINAL

COMPLETION ESTIMATE