Convergency of $\sum_{n=0}^{\infty}\frac{n!}{(kn)!}$, where $k > 1$ I am confident that $$\sum_{n=1}^{\infty}\frac{n!}{(2n)!}\approx1.5923$$ converges. Other series such as $$\sum_{n=1}^{\infty}\frac{n!}{(1.1n)!}\approx5.5690$$ appear to converge as well, and my hypothesis is that the series $$\sum_{n=0}^{\infty}\frac{n!}{(kn)!}, k > 1$$ converges. I know that the $p$-series converges for $p>1$ due to the following relationship: $$\int_{1}^{\infty}\frac{1}{x^p}dx < \sum_{1}^{\infty}\frac{1}{x^p}<1+\int_{1}^{\infty}\frac{1}{x^p}dx$$ However, I am not sure how to prove that $$\sum_{n=0}^{\infty}\frac{n!}{(kn)!}, k > 1$$ converges with the same method that is used to prove the convergency of the $p$-series. Could you provide me some hints?

$\log\Gamma(s+1)$ is a convex function on $\mathbb{R}^+$ and $$ \frac{n!}{(kn)!} = \frac{\Gamma(n+1)}{\Gamma(kn+1)}\leq \frac{1}{n^{(k-1)n}}.$$ Something similar but more accurate can be deduced from Stirling's inequality $$ \left(\frac{m}{e}\right)^m\sqrt{2\pi m}\, e^{\frac{1}{12m+1}}\leq\Gamma(m+1) \leq \left(\frac{m}{e}\right)^m\sqrt{2\pi m}\, e^{\frac{1}{12m}}.$$