How find this limit $\lim_{n\to \infty} \left(\frac{(2n)!}{2^n\cdot n!}\right)^{\frac{1}{n}}\cdot \cdots$ Find this limit
$$\lim_{n\to\infty}\left(\dfrac{(2n)!}{2^n\cdot n!}\right)^{\frac{1}{n}}\left(\tan{\left(\dfrac{\pi\sqrt[n+1]{(n+1)!}}{4\sqrt[n]{n!}}\right)}-1\right)$$
I know we must use this 
$$n!\approx\left(\dfrac{n}{e}\right)^{n}\sqrt{2n\pi}$$
so
$$\dfrac{\sqrt[n+1]{(n+1)!}}{\sqrt[n]{n!}}\approx\dfrac{n+1}{n}\dfrac{\sqrt[n+1]{2(n+1)\pi}}{\sqrt[n]{n!}}\to 1,n\to \infty$$
But I can't .Thank you
 A: $$L=\lim_{n\to\infty}\left(\dfrac{(2n)!}{2^n\cdot n!}\right)^{\frac{1}{n}}\left(\tan{\left(\dfrac{\pi\sqrt[n+1]{(n+1)!}}{4\sqrt[n]{n!}}\right)}-1\right)=\lim_{n\to\infty}\left(\dfrac{(2n)!}{2^n\cdot (n!)^2}\right)^{\frac{1}{n}}\cdot\lim_{n\to\infty}\sqrt[n]{n!}\left(\tan{\left(\dfrac{\pi\sqrt[n+1]{(n+1)!}}{4\sqrt[n]{n!}}\right)}-1\right)= \lim_{n\to\infty}a_n\cdot\lim_{n\to\infty}b_n$$
1) Apply root criterion (Chauchy-d'Alembert): $$ \lim_{n\to\infty} a_n = \lim_{n\to\infty}\frac{(2n+2)!}{2^{n+1}\cdot((n+1)!)^2}\cdot\frac{2^n\cdot(n!)^2}{(2n)!} =2$$
2) Note $\dfrac{\sqrt[n+1]{(n+1)!}}{\sqrt[n]{n!}}=c_n$$$\lim_{n\to\infty}b_n= \lim_{n\to\infty}\sqrt[n]{n!}\cdot(\tan(\frac{\pi}{4}c_n)-1)= \lim_{n\to\infty}\sqrt[n]{n!}\cdot\frac{\sqrt{2}\sin\frac{\pi}{4}(c_n-1)}{\cos\frac{\pi}{4}c_n}=\lim_{n\to\infty}\sqrt[n]{n!}\cdot\frac{2\sin\frac{\pi}{4}(c_n-1)}{\frac{\pi}{4}(c_n-1)}\cdot\frac{\pi}{4}(c_n-1)=\frac{\pi}{2}\lim_{n\to\infty}\sqrt[n]{n!}\cdot(c_n-1)= \frac{\pi}{2}\cdot\lim_{n\to\infty}(\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}) =\frac{\pi}{2}\cdot\frac{1}{e} = \frac{\pi}{2e}.$$
Applied $$\lim_{n\to\infty}(\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}) = \frac{1}{e}$$ the Lalescu's Sequence.(See http://www.artofproblemsolving.com/blog/44744 ) 
Conclusion $$L=  \frac{\pi}{e}.$$
A: Replace all factorials with their Stirling approximation, and apply l'Hopital's rule as many times as necessary. The end result should be $\dfrac\pi e$
