How do you do this question?

k > 1, or k ≥ 2

Step-by-step 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.

the answer is ! ?