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

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 $$a_n >\frac{n^2} {e^2}$$
