$\lim \limits_{n \to \infty\ }\sqrt[n^n]{(3n)!+n^n}$

Question

$\lim \limits_{n \to \infty\ }\sqrt[n^n]{(3n)!+n^n}$

$\sqrt[n^n]{(3n)!}\le\sqrt[n^n]{(3n)!+n^n}\le\sqrt[n^n]{(3n)!+(3n)!}=\sqrt[n^n]{2\cdot(3n)!}=\sqrt[n^n]{2}\cdot \sqrt[n^n]{(3n)!}$

But I haven't any idea what can I do with $\sqrt[n^n]{(3n)!}$

Answer 1

Just use the fact that $(3n)! \leq (3n)^{3n}$. This gives $(3n)!+n^{n} \leq 2(3n)^{3n}$. The answer is $1$.

Answer 2

We have that by Stirling $\log k! =k\log k-k+O(\log k)$

then

$$\log (\sqrt[n^n]{(3n)!})=\frac{\log ((3n)!)}{n^n}=\frac{(3n)\log ((3n))-3n+O(\log(3n))}{n^n}\to 0$$

therefore

$$\sqrt[n^n]{(3n)!}\to 1$$