I tried using $\liminf_{n\to\infty} \frac{a_{n+1}}{a_n} \leq \liminf_{n\to\infty} \sqrt[n]{a_n} \leq \limsup_{n\to\infty} \sqrt[n]{a_n} \leq \limsup_{n\to\infty} \frac{a_{n+1}}{a_n}$ for $a_n\geq 0$.
So i started like this:
$$\frac{\frac{n+1}{\sqrt[n+1]{(n+1)!}}}{\frac{n}{\sqrt[n]{n!}}}=\frac{(n+1)\cdot \sqrt[n]{n!}}{\sqrt[n+1]{(n+1)!}\cdot n}=\frac{(n+1)\cdot \sqrt[n]{n!}}{\sqrt[n]{n!(n+1)}\cdot \sqrt{n!(n+1)}\cdot n}=\frac{(n+1)\cdot \sqrt[n]{n!}}{\sqrt[n]{n!}\cdot \sqrt[n]{n+1}\cdot \sqrt{n!}\cdot \sqrt{n+1}\cdot n}=\frac{n+1}{\sqrt[n]{n+1}\cdot \sqrt{n!}\cdot \sqrt{n+1}\cdot n}$$ I don't see how this would help me - any ideas?
