Proof of $\sup\{ (n!)^{(\frac{1}{n})}| n ∈ N\} = +∞?$

Question

How do I prove if $\sup\{ (n!)^{1/n} | n \in\Bbb N\} = +\infty$?

I don't know how to properly explain it

Answer 1

Hint: suppose for simplicity that $n$ is even. Then $$n! = 1 \cdot 2 \cdots \frac{n}{2} \cdot \left(\frac{n}{2}+1\right) \cdots n \ge \left(\frac{n}{2}+1\right) \cdots n > \left(\frac{n}{2}\right)^{n/2}.$$ What can you conclude about $(n!)^{1/n}$?

Answer 2

For $k=1,2...n$ we have that $(k-1)(n-k) \geq 0.$

Thus $(n-k+1)k \geq n$

$$n \cdot 1 \geq n$$ $$(n-1) \cdot 2 \geq n$$ $$(n-2) \cdot 3 \geq n$$ $$.$$ $$.$$ $$.$$ $$.$$ $$.$$ $$2 \cdot (n-1) \geq n$$ $$1 \cdot n \geq n$$

Multiplying all the inequalities we have that $(n!)^2 \geq n^n$

Continue it from here..

Answer 3

From $(1+1/n)^n$ being a bounded increasing sequence with limit traditionally called $e$, you can show by induction that $n! > (n/e)^n$ so that $(n!)^{1/n} > n/e$.

Actually, $\lim_{n\to\infty} \dfrac{(n!)^{1/n}}{n} =\dfrac1{e} $.