Convergence of infinite sum (factorial and root)

Question

Find the convergence of the following sequence: $$\sum_{n=1}^{\infty}\left(\dfrac{n!}{(2n)!}\right)^{\frac{1}{n}}$$ I tried to use the ratio test. It wasn't useful. Also I used the logharitmic test (I'm not sure this is the standard name for it): $\lim_{n\to\infty}\dfrac{\ln\frac{1}{u_n}}{\ln n}=l$. If $l>1$, $\sum_{n=1}^{\infty}u_n$ converges. It didn't work to well. Thanks for your help!

Answer 1

Using Stirling's approximation, you get $$a_n=\left(\dfrac{n!}{(2n)!}\right)^{\frac{1}{n}} \sim \left( \frac{\sqrt{2 \pi n}}{\sqrt{4 \pi n}}\left(\frac{n}{e}\right)^n \left(\frac{e}{2n}\right)^{2n}\right)^{1/n}=\left(\frac{1}{\sqrt{2}}\right)^{1/n} \frac{e}{2n}$$

Hence the series diverges.

Answer 2

$$\left(\frac{n!}{(2n)!}\right)^{1/n}=\left(\frac1{(n+1)(n+2)\cdots(2n)}\right)^{1/n}>\left(\frac1{(2n)(2n)\cdots(2n)}\right)^{1/n}=\frac1{2n}$$

hence the series diverges.