Theorem: convergence for the ratio test implies convergence for the root test. So whenever the ratio test works (i.e. tells you whether the series converges), the root test also works and the limits coincide.
If $$\lim_{n\to\infty}\frac{|a_{n+1}|}{|a_n|}=L,$$ then $$\lim_{n\to\infty}|a_n|^{1/n}=L.$$
Question: a series that cannot be evaluated through the root test because algebra won't allow it (usually involving a factorial), but can be evaluated through the ratio test will it theoretically have the same limit value through root test (if it were possible to be evaluated) as ratio test due to the theorem?
Example: $$\displaystyle\sum\frac{n!}{n^n}$$
ratio test yields $ L = \frac{1}{e} < 1$
but if you tried the root test, you would be miserable. The limit is 1/e by ratio test implies root test?
(though you can just use Stirling's)
$$\frac{\sqrt{2\pi n} e^{-n} n^n}{n^n} \\ = \sqrt{2\pi} \cdot \frac{1}{e^n} \cdot \sqrt{n}$$
The limit $1 \le \sqrt{n}^{1/n} \le n^{1/n} \to 1$ as $n \to +\infty$ allows us to recover the ratio $1/e$
$$L = \lim\limits_{n\to+\infty} a_n^{1/n} = \lim\limits_{n\to+\infty} \frac1e \sqrt{n}^{1/n} = \frac1e$$
