I presume that the series is ∑(n!)ᵏ/(kn)!. Note that 0!=1.
When k=0, this becomes ∑(n!)⁰/0!=∑1/1=n+1→∞ as n→∞.
When k=1, it’s ∑n!/n!=n+1→∞ as n→∞.
So k>1 because x! is only defined for x≥0.
Let k=2, then we have ∑(n!)²/(2n)!=0!²/0!+1!²/2!+2!²/4!+3!²/6!+4!²/8!+...
That gives us:
1+1/2+1/6+1/20+1/70+1/252+...
For convergence the ratio between consecutive terms must be less than 1.
(1/6)/(1/2)=2/6=⅓=0.33<1;
(1/20)/(1/6)=6/20=0.3<1;
(1/70)/(1/20)=2/7=0.29<1;
(1/252)/(1/70)=70/252=5/18=0.28<1, so the ratio is decreasing for k=2.
Let’s check for k≥3 using general n.
(n+1)ᵏ/[(kn+k)(kn+k-1)...(kn+1)] is the formula for the ratio of consecutive terms, because the factorials cancel out. For example, if k=2, n=3:
4²/(8×7)=2/7; if n=4:
5²/(10×9)=5/18.
If k=3 and n=3:
4³/(12×11×10)=8/165=0.048<1.
Note that there are k multipliers in the denominator and the degree of n is also k, as in the numerator, making the denominator larger than the numerator. The factor k also occurs in the denominator and has degree k. As k increases the ratio therefore decreases. So convergence occurs for integer k≥2.