Evaluate $ \lim_{n\to \infty}\sqrt[n]{n!}$ I am trying to evaluate the $\lim(\sqrt[n]{n!})$ using 2 theorems (2 proofs)


Theorem 1: Let $\{c_n\}$ be any sequence in $\mathbb{R}^+$. Then, $\displaystyle \underline{\lim}\frac{c_{n+1}}{c_n}\leq \underline{\lim}\sqrt[n]{c_n}$ and $\displaystyle \overline{\lim}\sqrt[n]{c_n}\leq \overline{\lim}\frac{c_{n+1}}{c_n}$. 


so with 1. I have $\frac{(n+1)!}{n!}$ =$n+1$  which is $\overline{\lim}=\infty$


and 2 with $\sqrt[n]{n!}\geq\sqrt[n]{(n/2)^{n/2}}$=$\sqrt{\frac{n}{2}}$ which is $\overline{\lim}=\infty$
is it valid?

P.S I was not using theorem 1 right
 A: Without using the theorems, for this kind of problems which involve factorials, a very useful trick is Stirling approximation which write $$n!\approx n^n \sqrt{2\pi n}e^{-n}$$ that is to say $$\log(n!)\approx n\log(n)-n+\frac 12\log(2\pi n)$$ So, $$\frac 1n\log(n!)\approx \log(n)-1-\frac 12\frac{\log(2\pi n)}{n} $$ $$\sqrt[n]{n!}\approx e^{\log(n)-1}=\frac ne$$
A: Using an integral test for convergence, you can notice that $$\int_1^n \ln(x) dx \leq \ln(n!) = \sum\limits_{k=2}^n \ln(k) \leq \int_2^{n+1} \ln(x)dx.$$
Therefore, it can be deduced that $$e^{k_1(n)} \cdot e^{\ln(n)-1} \leq \sqrt[n]{n!} \leq e^{\ln(n)-1} \cdot e^{k_2(n)},$$ for some $k_1(n),k_2(n) \to 0$, hence $$\sqrt[n]{n!} \underset{n \to + \infty}{\sim} n \cdot e^{-1}.$$
So, indeed $\sqrt[n]{n!} \to + \infty$, but we also know that $\frac{\sqrt[n]{n!}}{n} \to e^{-1}$.
A: Note that
$$(2n)! > \prod_{k=n}^{2n}k >n^{n+1}$$
and
$$[(2n)!]^{1/2n} > n^{1/2n}n^{1/2}>\sqrt{n}.$$
Similarly show
$$[(2n+1)!]^{1/(2n+1)} >\sqrt{n}.$$
Hence,
$$(n!)^{1/n} > \sqrt{\left \lfloor{n/2}\right \rfloor}\rightarrow \infty$$
A: Using Stolz–Cesàro:
$$
\log L = \lim_{n\to\infty}\log(\sqrt[n]{n!}) = \lim_{n\to\infty}\frac{\log1+\cdots+\log n}n=
\lim_{n\to\infty}\frac{\log(n+1)}1=\infty,
$$
so,
$$\lim_{n\to\infty}\log(\sqrt[n]{n!}) = \infty.$$
