Evaluate $ \lim_{n\to \infty}\sqrt[n]{n!}$

Question

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

Answer 1

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, $$\frac 1n\log(n!)\approx \log(n)-1-\frac 12\frac{\log(2\pi n)}{n} $$ $$\sqrt[n]{n!}\approx e^{\log(n)-1}=\frac ne$$

Answer 2

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,

$$(n!)^{1/n} > \sqrt{\left \lfloor{n/2}\right \rfloor}\rightarrow \infty$$

Answer 3

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 $\frac{\sqrt[n]{n!}}{n} \to e^{-1}$.

Answer 4

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, $$\lim_{n\to\infty}\log(\sqrt[n]{n!}) = \infty.$$