This question already has an answer here:
The answer to your question is yes: for any $L \in [0,\infty]$, if $\frac{a_{n+1}}{a_n} \rightarrow L$, then also $a_n^{\frac{1}{n}} \rightarrow L$.
Here is a stronger result:
Theorem: For any sequence $\{a_n\}_{n=1}^{\infty}$ of positive real numbers one has
$$\liminf_{n \rightarrow \infty} \frac{a_{n+1}}{a_n} \leq \liminf_{n \rightarrow \infty}
a_n^{\frac{1}{n}} \leq \limsup_{n \rightarrow \infty} a_n^{\frac{1}{n}} \leq
\limsup_{n \rightarrow \infty} \frac{a_{n+1}}{a_n}$$.
For a proof see e.g. $\S 5.3$ of these notes.
Note that the converse is not true: it is possible for the root test limit to exist but the ratio test limit not to. In fact we can get this just by mildly rearranging the terms of a convergent geometric sequence,e.g.
$\frac{1}{2}, \frac{1}{8}, \frac{1}{4}, \frac{1}{32}, \frac{1}{16} \ldots$
Here half of the successive ratios are $\frac{1}{4}$ and the other half are $2$, so the ratio test limit does not exist. But the root test limit is still $\frac{1}{2}$.
The limit
$\lim_{n\to\infty}\sup\left|\frac{a_{n+1}}{a_n} \right| = L < 1$
is always greater than or equal to the limit
$\lim_{n\to\infty}\sup\left| a_n^{1/n} \right| = L$
So the root test is stronger than the ratio test. One can find cases when the root test shows convergence but the ratio test does not.
In fact, the ratio test is a corollary of the root test. For example see: S. Krantz Real Analysis and Foundations, Chapman and Hall/CRC (Remark 4.1 on page 105, second edition of the book). It comes out directly from the proof of the ratio test.
Yes，you can show that by proving the following:
$$
if\quad a_n\ge 0\quad and\quad \lim_{n\to\infty}a_n=L,\quad then
$$
$$
\lim_{n\to\infty}(\prod_{k=1}^{k=n} a_k)^{\frac{1}{n}}=L
$$
Check by definition