gpt 6 astra ultra attempt on Math-290

Irrationality of Euler's Constant

Definitions and mathematical statement

Euler's constant is defined using natural logarithms by \[ \gamma=\lim_{n\to\infty}\left(H_n-\log n\right), \qquad H_n=\sum_{j=1}^n\frac1j. \] The selected question is whether $\gamma\notin{\mathbb{Q}}$, equivalently whether $q\gamma-p\ne0$ for every $p\in{\mathbb{Z}}$ and integer $q\ge1$. The sequence defining the limit, rather than a finite decimal expansion, specifies the constant exactly. This is not Euler's exponential constant $e$, and transcendence of $\gamma$ would be a stronger conclusion than the requested irrationality.

Formulation and concept references: [S1].

Short English statement

Is the limiting difference between the harmonic sum and the natural logarithm an irrational number?

Research attempt

Scope and literature check.

This attempt was prepared by GPT-6 Astra Ultra on 27 September 2026. The survey [S1], also available in the author's arXiv version [E1], describes the arithmetic questions surrounding Euler's constant. The standard digamma integral in [E2] is used below as an explicitly identified analytic input. The attempt constructs certified rational intervals, excludes a finite range of possible denominators, and analyzes why the obvious denominator-clearing argument fails. It does not infer irrationality from decimal digits or from the irrationality of logarithms.

All logarithms are natural. Set $a_n=H_n-\log n$ and $R_n=a_n-\gamma$. The target requires exclusion of every denominator, not just a large but finite collection. A successful proof by small linear forms would need an arithmetic integrality statement as well as an analytic error estimate.

Elementary signed remainder bounds.

The consecutive difference is \[ a_k-a_{k+1}=\log(1+1/k)-\frac1{k+1} =\int_k^{k+1}\frac{k+1-t}{t(k+1)}\,dt>0. \] Summing to infinity yields $R_n>0$. On the integration interval, \[ \frac1{2(k+1)^2} <\log(1+1/k)-\frac1{k+1} <\frac1{2k(k+1)}. \] The upper sum telescopes; the lower sum exceeds the corresponding decreasing-function integral. Therefore \[ \frac1{2(n+1)}<R_n<\frac1{2n}. \tag{1} \] In particular \[ H_n-\log n-\frac1{2n}<\gamma <H_n-\log n-\frac1{2(n+1)}. \tag{2} \] The signs in these bounds are part of the argument. A bound on $|R_n|$ alone would not supply the nonzero linear forms needed in an irrationality proof.

A sharper integral remainder with a controlled sign.

The digamma identities [E2], valid at every positive integer $n$, give \[ R_n=\frac1{2n} -2\int_0^\infty\frac{t}{(n^2+t^2)(e^{2\pi t}-1)}\,dt. \tag{3} \] Specifically $\psi(n)=H_n-1/n-\gamma$ and the integral formula for $\psi(n)$ imply $3$. We are citing this identity, not claiming to derive the theory of the gamma function here.

Now expand the rational factor by the finite geometric identity \[ \frac1{n^2+t^2} =\sum_{j=0}^{m-1}\frac{(-1)^jt^{2j}}{n^{2j+2}} +\frac{(-1)^mt^{2m}}{n^{2m}(n^2+t^2)}. \tag{4} \] The remainder has a fixed sign for $t>0$. Its magnitude is strictly less than the first omitted monomial. The moments are \[ 2\int_0^\infty\frac{t^{2k-1}}{e^{2\pi t}-1}\,dt =\frac{2(2k-1)!\zeta(2k)}{(2\pi)^{2k}} =\frac{|B_{2k}|}{2k}. \] The first equality follows by the positive geometric series for the denominator and termwise integration; the second is the classical even-zeta evaluation in Bernoulli numbers. Thus, with \[ A_{n,m}=H_n-\frac1{2n} +\sum_{k=1}^m\frac{B_{2k}}{2k n^{2k}}, \] we have \[ \gamma=A_{n,m}-\log n+\varepsilon_{n,m},\quad 0<(-1)^m\varepsilon_{n,m} <\frac{|B_{2m+2}|}{(2m+2)n^{2m+2}}. \tag{5} \] For $m=4$, the correction is positive. It is important that $4$ is a finite identity with an integral remainder, rather than an assertion that the full asymptotic series converges for fixed $n$.

Replacing the logarithm by a rational interval.

For $n>1$ put $z=(n-1)/(n+1)\in(0,1)$. Integrating the geometric series for $1/(1-z^2)$ gives \[ \log n=2\sum_{j=0}^{\infty}\frac{z^{2j+1}}{2j+1}. \] Let $L_{n,K}$ denote the sum through $j=K-1$. Its positive tail satisfies \[ 0<\log n-L_{n,K} <T_{n,K}:=\frac{2z^{2K+1}}{(2K+1)(1-z^2)}. \tag{6} \] Every quantity in $6$, except the logarithm itself, is rational and can be evaluated exactly. Combining $5$ at $m=4$ with $6$ yields \[ A_{n,4}-L_{n,K}-T_{n,K}<\gamma <A_{n,4}-L_{n,K}+\frac1{132n^{10}}. \tag{7} \] Here \[ A_{n,4}=H_n-\frac1{2n}+\frac1{12n^2}-\frac1{120n^4} +\frac1{252n^6}-\frac1{240n^8}, \] and $B_{10}/10=1/132$ fixes the last coefficient. The lower endpoint uses the upper bound on $\log n$; the upper endpoint uses its lower bound. Reversing that subtraction would destroy the certificate.

A finite denominator exclusion, stated at its actual scope.

Take $n=32$, $K=400$, and denote the two rational endpoints in $7$ by $L<U$. Their width is less than $7\cdot10^{-18}$, checked by rational cross multiplication. The local script checks every integer $1\le q\le100000$. For each $q$, the smallest possible numerator of a rational strictly above $L$ is \[ p_q=\lfloor qL\rfloor+1. \] It verifies $p_q/q\ge U$ for all those $q$. Thus the certified interval contains no rational with positive denominator at most $100000$. Reduction to lowest terms is unnecessary: testing every denominator also includes unreduced representations.

This is an exact finite conclusion, not a comparison with a floating-point value of $\gamma$. The decimal width is only a readable summary; the script uses fractions throughout. The result remains consistent with $\gamma$ being rational with a larger denominator. It is not asserted as a record denominator bound.

The simplest clearing attempt diverges.

Let $d_n=\operatorname{lcm}(1,\ldots,n)$, so $d_nH_n$ is an integer. One might try to multiply the defining approximation by $d_n$ and use its small remainder. Two separate obstructions appear. First, \[ d_n(H_n-\gamma-\log n)=d_nR_n \] still contains $d_n\log n$, which is not an integer and has not been arithmetically eliminated. Second, even its positive size does not tend to zero. Since $n$ and $n-1$ are coprime and both divide $d_n$, \[ d_n\ge n(n-1),\qquad d_nR_n>\frac{n(n-1)}{2(n+1)}\longrightarrow\infty. \tag{8} \] This elementary lower bound already kills that route; no prime number theorem estimate for $d_n$ is needed.

Using more terms in $5$ improves the analytic remainder, but creates additional denominators from powers of $n$ and Bernoulli numbers. Using $6$ removes the explicit logarithm only by introducing still more rational denominators. A valid irrationality argument must estimate their combined growth, rather than count the number of matching digits.

An abstract criterion and an adversarial test.

Suppose integers $A_j,B_j$, with $B_j>0$, could be constructed such that \[ 0<|B_j\gamma-A_j|\longrightarrow0.\tag{9} \] Then $\gamma$ would be irrational. Indeed, if $\gamma=p/q$, the nonzero integer $B_jp-qA_j$ would eventually have absolute value less than one. Formula $9$ is stronger than saying rational approximants $A_j/B_j$ converge quickly; it controls the error after multiplying by the denominator and requires nonvanishing.

For a simple countertest to the weaker claim, take the rational target $c=0$ and approximants $r_j=1/j!$. Their errors decay faster than any fixed geometric rate, but with denominator $B_j=j!$ the nonzero scaled error is always one. Fast convergence by itself is therefore compatible with a rational limit. Likewise the finite exclusion above does not construct a sequence satisfying $9$.

Verification and final status.

The proof of $1$ is elementary and independent of the digamma input. The signs and first omitted coefficient in $5$ are checked using $4$. The interval and the denominator exclusions are reproducible exact arithmetic. The source identity $3$ and the even-zeta moments are stated as external standard inputs with their positive-real hypotheses.

UNRESOLVED. The strongest arithmetic conclusion proved here is exclusion of denominators at most $100000$ by an explicit rational certificate, together with the obstruction $8$ to naive clearing. The first missing step toward irrationality is an infinite family of nonzero integer linear forms as in $9$, with proved denominator control. Specific next targets are simultaneous cancellation of logarithmic terms, denominator estimates for an accelerated integral construction, and a nonvanishing argument surviving that cancellation. None has been established in this attempt.

Sources

• S1 Jeffrey C. Lagarias, "Euler's constant: Euler's work and modern developments," Bull. Amer. Math. Soc. 50 $2013$, no. 4, 527-628. https://www.ams.org/journals/bull/2013-50-04/S0273-0979-2013-01423-X/S0273-0979-2013-01423-X.pdf. Formulation source.
• E1 Jeffrey C. Lagarias, Euler's constant: Euler's work and modern developments, author preprint of [S1]. https://arxiv.org/abs/1303.1856.
• E2 NIST Digital Library of Mathematical Functions, Section 5.9(ii), especially equations 5.9.15 and 5.9.16. https://dlmf.nist.gov/5.9#E15. Explicit analytic input for the signed integral remainder.