For a non-negative real series $(a_n)_{n \in \mathbb{N}}$, the tests give two (possibly undefined) numbers: let’s call them $L_\textit{root} := \lim_n (a_n)^{\frac{1}{n}}$, and $L_\textit{ratio} := \lim_n \frac{a_{n+1}}{a_n}$.
From Lemma 3 of these notes by Pete L. Clark, it follows that if $L_{\textit{ratio}}$ is defined, then $L_\textit{root}$ is also defined, and they are equal.
This is reasonably intuitive, with a bit of thought: suppose that for $n>N$, the ratio of consecutive terms $\frac{a_{n+1}}{a_n}$ is always close to $L$. Then (still for n>N), consider $a_n$ as produced by multiplying $a_N$ by all the later consecutive ratios; so it’s close to $L^{n-N} a_N$, and its $n$th root is close to $(L^{n-N} a_N)^{\frac{1}{n}} = L (\frac{a_N}{L^N})^\frac{1}{n}$. The second factor here, being the $n$th root of a constant, goes to $1$ as $n$ grows; so for sufficiently large $n$, $(a_n)^\frac{1}{n}$ will be close to $L$. (Exercise: make this argument precise — replace each “…close to…” by appropriate specific bounds.)
On the other hand, the converse doesn’t generally hold. $L_\textit{root}$ may be defined even if $L_\textit{ratio}$ is not. For example, set $a_n = 2^n$ when $n$ is even, $a_n = 2^{n-1}$ when $n$ is odd. Then the ratio of consecutive terms alternates between 1 and 4, so $L_\textit{ratio}$ is undefined; but the sequence is close enough to $2^n$ that the root converges, with $L_\textit{root} = 2$.
(Thanks to @David Mitra’s comment for the reference to the linked notes.)