Let $ \{a_k\} $ be a sequence of positive real numbers. Why does it hold that
$$ \liminf \frac{a_{k+1}}{a_{k}} \leq \liminf (a_k)^{\frac{1}{k}}\leq\limsup (a_k)^{\frac{1}{k}} \leq \limsup \frac{a_{k +1}}{a_ {k}} $$
This question has arisen from me trying to understand why
$$ \limsup \left| \frac{a_{k+1}}{a_k}\right| = R \implies \frac{1}{\limsup |a_k|^{1/k}} = R $$
If $\liminf \frac{a_{k+1}}{a_k} = 0$, the first inequality is clear. Thus let us suppose it is strictly positive.
Let $0 < c < \liminf \frac{a_{k+1}}{a_k}$. Then there is a $K_c$ such that for all $k \geqslant K_c$ we have $\frac{a_{k+1}}{a_k} > c$, and hence
$$a_{K_c+n} = a_{K_c}\prod_{j=0}^{n-1} \frac{a_{K_c+j+1}}{a_{K_c+j}} > a_{K_c}\cdot c^n = \frac{a_{K_c}}{c^{K_c}}\cdot c^{K_c+n},$$
which implies
$$\left(a_{K_c+n}\right)^{1/(K_c+n)} > \underbrace{\left(\frac{a_{K_c}}{c^{K_c}}\right)^{1/(K_c+n)}}_{\xrightarrow[n\to\infty]{} 1}\cdot c$$
and therefore $\liminf (a_k)^{1/k} \geqslant c$. Since that holds for all $0 < c < \liminf \frac{a_{k+1}}{a_k}$, the inequality
$$\liminf \frac{a_{k+1}}{a_k} \leqslant \liminf a_k^{1/k}$$
follows. The third inequality is seen analogously, and the second, $\liminf a_k^{1/k} \leqslant \limsup a_k^{1/k}$ is a special case of the general inequality between the limites inferiores and superiores.
The purported equality
$$\limsup \left| \frac{a_{k+1}}{a_k}\right| = R \implies \frac{1}{\limsup |a_k|^{1/k}} = R$$
is however generally false. Consider $$a_k = \begin{cases}1 &, k \text{ odd}\\ 2 &, k \text{ even}\end{cases}$$ for a simple counterexample. $\limsup \frac{a_{k+1}}{a_k} = 2$, but $a_k^{1/k} \to 1$.
If $\lim \left\lvert\frac{a_{k+1}}{a_k}\right\rvert$ exists, then we have
$$\lim \left\lvert\frac{a_{k+1}}{a_k}\right\rvert = \lim \lvert a_k\rvert^{1/k},$$
however.
These inequalities are addressed in $\S$ 11.5.3 of these notes. (More precisely the middle inequality is clear, and the two outer inequalities are very similar, so the notes carefully prove one of them.)
As Daniel Fischer has pointed out, the implication you ask about at the end of your question is false in general. What is true is $\lim_{n \rightarrow \infty} |\frac{a_{n+1}}{a_n}| = R \implies \lim_{n \rightarrow \infty} |a_n|^{\frac{1}{n}} = R$ ("The root test is stronger than the ratio test") and this follows immediately from the above inequalities.
You mention "radius of convergence" in your title but not in the body of your question. Maybe you meant to ask about it. If so, the Cauchy-Hadamard Theorem says that the radius of convergence of the (real or complex) series $\sum_{n=0}^{\infty} a_n x^n$ is $\frac{1}{\overline{\theta}}$, where $\overline{\theta} = \limsup_n |a_n|^{\frac{1}{n}}$. For this see $\S$ 11.8.1 of the same set of notes.
