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

Edit

After @Gary's comment, if you want batter, use the full Stirling expansion $$\log(n!)=n (\log (n)-1)+\frac{1}{2} (\log (n)+\log (2 \pi ))+\frac{1}{12n} +O\left(\frac{1}{n^2}\right)$$ Divide by $n$ and exponentiate $$\sqrt[n]{n!}=\frac n e +\frac {\log (2 \pi n)} {2e} +O\left(\frac{\log^2(n)}{n}\right)$$

Answer 2

Elementary answer:

In the expression of $n!$, at least $\frac n2$ factors are above $\frac n2$ so that $n!>\left(\frac n2\right)^{n/2}$.

Hence $\sqrt[n]{n!}>\sqrt{\frac n2}$ and the limit diverges.

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

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 5

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

Answer 6

As an alternative we can use that bound

$$n! \ge \frac{n^n}{e^n}$$

therefore

$$\sqrt[n]{n!}\ge \frac n e\to \infty$$

Answer 7

Proposition. $\lim_{n\to\infty}{\sqrt[n]{n!}}=+\infty$.

Proof. Take an arbitrary $Μ>0$. Thus, since $\frac{M^{n+1}}{(n+1)!}\Big/\frac{M^n}{n!} =\frac{M}{n+1}\to 0<1$, by the Ratio Test we have that $\lim_{n\to\infty}{\frac{M^{n}}{n!}}=0$, which means that there exists a natural number $N\geq 1$ such that $\frac{M^n}{n!}<1$ for all $n\geq N$, or equivalently, $\sqrt[n]{n!}>M$ for all $n\geq N$ and we're done.

Answer 8

Another way, by ratio-root criterion

$$\frac{a_{n+1}}{a_n} \rightarrow L\implies a_n^{\frac{1}{n}} \rightarrow L$$

in this case by $a_n=n!$ we have

$$\frac{a_{n+1}}{a_n}=\frac{(n+1)!}{n!}=n+1\to \infty \implies \left(a_n^{\frac{1}{n}}\right)=\sqrt[n]{n!} \to \infty$$

Answer 9

From the series for $e^x$,

$\begin{array}\\ e^n &=\sum_{k=0}^{\infty} \dfrac{n^k}{k!}\\ &>\dfrac{n^n}{n!} \qquad \text{Chosing term with } k=n\\ \end{array} $

so $n! \gt \dfrac{n^n}{e^n} =(\frac{n}{e})^n $ so $(n!)^{1/n} \gt \frac{n}{e} \to \infty$.