For integers d≥2 and a>b≥1, put Cd,n(a,b)=(a^dⁿ⁺¹-b^dⁿ⁺¹)/(adⁿ-bdⁿ) and Sd(a,b)=∑n≥0Cd,n(a,b)⁻¹. Let qN be the denominator in lowest terms of the partial sum ∑ₙ₌₀NCd,n(a,b)⁻¹. We obtain an exact structural description of qN by evaluating the natural common denominator QN and proving that the cancellation factor QN/qN is eventually periodic. Writing μ(ξ) for the irrationality exponent of ξ, we obtain\[μ( S_d(a,b))=d(d-1)log a/dlog a-log(a,b)(d≥4).\]For $d=3$, the same formula holds when log(a,b)/log a≥3(2-√3), while μ( S₃(a,1))=2 for every a≥2. For every fixed integer m≥1 and every periodic sequence (ωₙ) of nonzero integers, the factor cancelled from QNᵐ in the partial sums of ∑n≥0ωₙ Cd,n(a,b)⁻ᵐ is also eventually periodic; for d≥4, the corresponding weighted sum has the irrationality exponent displayed above. For $d=2,3$ and $b=1$, the exponent is $2$ whenever m is a power of $2$ and the weights have a common $2$-adic valuation. Let (Uᵣ) be a first-kind Lucas sequence with coprime integer parameters $P,Q$, where Q≠0, whose real characteristic roots satisfy α>1 and |β|<α. Let h≥1 be an integer. For\[∑n≥0ω_n(h(d-1)d^n{U_{dⁿ⁺¹}}{Ud^n})⁻ᵐ,\]we determine the reduced partial-sum denominators, and the cancelled factors are again eventually periodic. For d≥4, its irrationality exponent is d(d-1)log(hα)/(dlogα+(d-1)log h). All three families are transcendental for every d≥2.
No takes yet. Share an insight, caveat, or question.
Wei Xie (2026) studied this question.
Synapse has enriched 5 closely related papers on similar clinical questions. Consider them for comparative context: