From the ratio to the n-th root

Proof. Let $w_n=\ln a_n$. Then $w_{n+1}-w_n=\ln\frac{a_{n+1}}{a_n}\to\ln L$. By the Cesàro lemma (part 2), $\frac{w_n}n\to\ln L$, so $a_n^{1/n}=e^{w_n/n}\to L$.

First limit. Let $a_n=\frac{n!}{n^n}$. Then
$$\frac{a_{n+1}}{a_n}=\frac{(n+1)\,n^n}{(n+1)^{n+1}}=\left(1+\frac1n\right)^{-n}\to\frac1e.$$
So $a_n^{1/n}=\frac{(n!)^{1/n}}n\to\frac1e$.

Second limit. For $a_n=\binom{2n}n$,
$$\frac{a_{n+1}}{a_n}=\frac{(2n+2)(2n+1)}{(n+1)^2}=\frac{2(2n+1)}{n+1}\to4.$$
So $\binom{2n}n^{1/n}\to4$.

Numerics. For $n=50$: $\frac{(50!)^{1/50}}{50}=0.3897$, close to $\frac1e=0.3679$.

Remark. The converse is false in general (the ratio can oscillate while the root converges), and this is why the root test for series is more powerful than the ratio test.