Answer
Hence, the series converges for $k\geq 2$.
Work Step by Step
$\lim\limits_{n \to \infty}|\dfrac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\frac{\dfrac{(n+1!)^{2}}{(k(n+1))!}}{\dfrac{(n!)^{2}}{(kn)!}}|$
$=\lim\limits_{n \to \infty}\frac{(n+1)^{2}}{(kn+1)(kn+2)....(kn+k)}$
When $k=1$
$\lim\limits_{n \to \infty}\frac{(n+1)^{2}}{(kn+1)(kn+2)....(kn+k)}=\lim\limits_{n \to \infty}\frac{(n+1)^{2}}{(n+1)} =\infty$
Thus, the series diverges when $k=1$
When $k=2$
$\lim\limits_{n \to \infty}\frac{(n+1)^{2}}{(kn+1)(kn+2)....(kn+k)}=\lim\limits_{n \to \infty}\frac{(n+1)^{2}}{(2n+1)(2n+2)} =\frac{1}{4}$
Thus, the series converges when $k=2$
When $k\gt 2$, we have a rational function with the numerator having a smaller degree than the denominator.
Thus, the limit must be $0$ when $n \to \infty $.
Hence, the series converges for $k\geq 2$
