This is what we have $$\lim_{n\rightarrow\infty} \frac{c^{n}}{n!^{\frac{1}{k}}},$$ $$n \in N, k>0, c>0$$ If n->inf $$\frac{{x}_{n+1}}{{x}_{n}}=\frac{{c}^{n+1}}{{(n+1)!}^{\frac{1}{k}}}*\frac{{n!}^{\frac{1}{k}}}{{c}^{n}}=\frac{c}{{(n+1)}^{\frac{1}{k}}}<1$$ obviously$${x}_{n}>0$$ This means, that we have a limit; let it be $$\lim_{n\rightarrow\infty} {x}_{n}=a$$ $${x}_{n+1}=\frac{c}{{(n+1)}^{\frac{1}{k}}}*{x}_{n}$$ $$a=0*a$$ this means that $$a=0$$
am i right?