This question is related to Examples 3.35 (a) and (b) in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition, p. 67.
Let us consider the series $$ \frac 1 2 + \frac 1 3 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{2^3} + \frac{1}{3^3} + \frac{1}{2^4} + \frac{1}{3^4} + \cdots$$ for which the formula for the general term $a_n$ is given by $$a_n = \begin{cases} \frac{1}{2^k} \ \mbox{ if } \ n = 2k-1 \\ \frac{1}{3^k} \ \mbox{ if } \ n = 2k \end{cases} $$ for $k = 1, 2, 3, \ldots$, and the series $$\frac 1 2 + 1 + \frac 1 8 + \frac 1 4 + \frac{1}{32} + \frac{1}{16} + \cdots$$ for which the formula for the general term is given by $$b_n = \begin{cases} \frac{1}{2^n} = \frac{1}{2^{2k-1}} \ \mbox{ if } \ n = 2k-1 \\ \frac{1}{2^{n-2}} = \frac{1}{2^{2k-2}} \ \mbox{ if } \ n = 2k \end{cases} $$ for $k = 1, 2, 3, \ldots$.
Now Rudin states that $$\liminf_{n\to\infty} \frac{a_{n+1}}{a_n} = \lim_{n\to\infty} \left( \frac 2 3 \right)^n = 0,$$ $$\liminf_{n\to\infty} \sqrt[n]{a_n} = \lim_{n\to\infty} \sqrt[2n]{\frac{1}{3^n}} = \frac{1}{\sqrt{3}},$$ $$\limsup_{n\to\infty} \sqrt[n]{a_n} = \sqrt[2n]{\frac{1}{2^n}} = \frac{1}{\sqrt{2}},$$ $$\limsup_{n\to\infty} \frac{a_{n+1}}{a_n} = \lim_{n\to\infty} \frac 1 2 \left( \frac 3 2 \right)^n = +\infty.$$ And, Rudin also states that $$ \liminf_{n\to\infty} \frac{b_{n+1}}{b_n} = \frac 1 8,$$ $$ \limsup_{n\to\infty} \frac{b_{n+1}}{b_n} = 2,$$ $$\lim_{n\to\infty} \sqrt[n]{b_n} = \frac 1 2.$$
How to rigorously verify these statements using machinery (i.e. the definitions and theorems ) developed by Rudin up to this point?
I know that $\liminf$ and $\limsup$ are the infimum and supremum, resp., of the set of all the subsequential limits (in the extended real number system) of a sequence, and there is a subsequence each converging to $\liminf$ and $\limsup$.
Moreover, for each $k \in \mathbb{N}$, we have $$\frac{a_{2k} }{a_{2k-1}} = \frac{ \frac{1}{3^k} }{ \frac{1}{2^k} } = \left( \frac 2 3 \right)^k \to 0 \ \mbox{ as } \ k \to \infty,$$ $$\frac{ a_{2k+1} }{ a_{2k} } = \frac{ \frac{1}{2^{k+1}}}{ \frac{1}{3^k} } =\frac 1 2 \left( \frac 3 2 \right)^k \to +\infty \ \mbox{ as } \ k \to \infty,$$ $$\sqrt[2k]{a_{2k}} = \sqrt[2k]{ \frac{1}{3^k}} = \sqrt{\frac 1 3} \to \sqrt{\frac 1 3} \ \mbox{ as } \ k \to \infty,$$ $$\sqrt[2k-1]{a_{2k-1}} = \sqrt[2k-1]{ \frac{1}{2^k}} = \frac{1}{2^{\frac{k}{2k-1}}} \to ? \ \mbox{ as } \ k \to \infty.$$ How to find $\lim_{k \to \infty} \sqrt[2k-1]{2^k} $ using what Rudin has established (in Theorem 3.20)?
Also, $$\frac{ b_{2k} }{ b_{2k-1} } = \frac{ \frac{1}{ 2^{2k-2} } }{ \frac{1}{ 2^{2k-1} } } = 2 \to 2 \ \mbox{ as } \ k \to \infty,$$ $$\frac{ b_{2k+1} }{ b_{2k} } = \frac{ \frac{1}{ 2^{2k+1} } }{ \frac{1}{ 2^{2k-2} } } = \frac 1 8 \to \frac 1 8 \ \mbox{ as } \ k \to \infty,$$
$$\sqrt[2k]{b_{2k}} = \sqrt[2k]{\frac{1}{2^{2k-2}}} = \frac{ \sqrt[2k]{4} }{ 2 } \to \frac 1 2 \ \mbox{ as } \ k \to \infty,$$ $$\sqrt[2k-1]{b_{2k-1}} = \sqrt[2k-1]{\frac{1}{2^{2k-1}}} = \frac 1 2 \to \frac 1 2 \ \mbox{ as } \ k \to \infty,$$
For part (b) notice that if $n$ is even:
$$\frac{a_{n+1}}{a_n}=1/8$$
and if $n$ is odd:
$$\frac{a_{n+1}}{a_n}=2$$
(as you state in your post).
Therefore the only subsequential limits of the sequence
$$\Bigg \{ \frac{a_{n+1}}{a_n} \Bigg \}$$
are $1/8$ and $2$.
i.e. any subsequence that converges either converges to $1/8$ or to $2$.
Therefore
$$\lim_{n \to \infty} \inf \frac{a_{n+1}}{a_n} = 1/8 $$
$$\lim_{n \to \infty} \sup \frac{a_{n+1}}{a_n} = 2 $$
Now if n is odd $\sqrt[n] a_n = 1/2$ and if n is even $\sqrt[n] a_n = \sqrt[n] 4 / 2$
$$\lim_{n \to \infty} \sqrt[n] 4 / 2 = \frac{1}{2} \lim_{n \to \infty} exp(\frac{1}{n} ln(4))$$
$$=\frac{1}{2} exp(ln(4) \lim_{n \to \infty} 1/n)$$
$$=\frac{1}{2}exp(0)$$
$$=1/2$$.
Therefore
$$\lim_{n \to \infty} \sqrt[n] a_n = 1/2$$