Answer:
k > 1, or k ≥ 2
Explanation:
aₙ = (n!)² / (kn)!
aₙ₊₁ = ((n+1)!)² / (k(n+1))!
lim(n→∞)│aₙ₊₁ / aₙ│
lim(n→∞)│[((n+1)!)² / (k(n+1))!] / [(n!)² / (kn)!]│
lim(n→∞)│[((n+1)!)² / (k(n+1))!] × [(kn)! / (n!)²]│
lim(n→∞)│[((n+1)!)² / (n!)²] × [(kn)! / (k(n+1))!]│
lim(n→∞)│[((n+1)! / n!)²] × [(kn)! / (kn+k)!]│
lim(n→∞) (n+1)² × (kn)! / (kn+k)!
This converges when the limit is less than 1.
If k=1:
lim(n→∞) (n+1)² × n! / (n+1)! = lim(n→∞) (n+1) = ∞
If k=2:
lim(n→∞) (n+1)² × (2n)! / (2n+2)! = lim(n→∞) (n+1)² / [(2n+2)(2n+1)] = 1/4
If k=3:
lim(n→∞) (n+1)² × (3n)! / (3n+3)! = lim(n→∞) (n+1)² / [(3n+3)(3n+2)(3n+1)] = 0
The series converges for k > 1, or k ≥ 2.