Analysis: Prove divergence of sequence $(n!)^{\frac2n}$

Question

I am trying to prove that the sequence $$a_n = (n!)^{\frac2n}$$ tends to infinity as $ n \to \infty $.

I've tried different methods but I haven't really got anywhere. Any solutions/hints?

Answer 1

Pick $K$ as large as you want, then $n!\ge K^{n-K}$ (for $n\ge K$), so $$(n!)^{1/n}\ge K^{(n-K)/n}=K^{1-K/n}\to K\text{ as }n\to\infty.$$

Answer 2

Hint: $$n!>\left(\frac{n}{2}\right )^n$$ for large enough $n$.

Answer 3

Note that $x(n+1-x)\ge (1)(n)$ for $1\le x\le n$. It follows that $$(n!)^2=(1)(n)(2)(n-1)(3)(n-2)\cdots (n)(1)\ge n^n$$ and therefore $(n!)^{2/n}\ge n$.

Answer 4

Let $c_n=(n!)^2$; then $\displaystyle\frac{c_{n+1}}{c_n}=\frac{((n+1)!)^2}{(n!)^2}=(n+1)^2\to\infty,$ $\;\;\;$so $(n!)^{\frac{2}{n}}=(c_n)^{\frac{1}{n}}\to\infty$

Answer 5

Let $$a_n = (n!)^{\frac2n}$$ $$\log(a_n)={\frac2n} \log(n!)$$and, as suggested by GPerez, use Stirling approximation, that is to say $$n!\approx n^n \sqrt{2\pi n}e^{-n}$$ or, even better, Stirling series $$\log(n!) \sim n\log(n)-n+\frac 12 \log(2\pi n)+\frac{1}{12n}+\cdots$$ So $$\log(a_n)\sim 2\frac{ n\log(n)-n+\frac 12 \log(2\pi n)+\cdots}{n}> 2(\log(n)-1) =\log\Big(\frac{n^2} {e^2}\Big)$$ and then $$a_n >\frac{n^2} {e^2}$$