$\lim \sup$ and $\lim \inf$ of root and ratio test sequences of a series (Rudin)
I understand that:
$$(1) \qquad \{\frac{a_{n+1}}{a_n}\} = \{\frac{2}{3},\frac{3}{2^2},\frac{2^2}{3^2},\frac{3^2}{2^3},...\} = \{\frac{2^n}{3^n}, \frac{3^n}{2^{n+1}}\}$$
$$(2) \qquad \{\sqrt[n]{a_n}\} = \{\sqrt[1]{1/2},\sqrt[2]{1/3},\sqrt[3]{1/2^2},\sqrt[4]{1/3^2},...\} = \{\sqrt[2n-1]{\frac{1}{2^n}}, \sqrt[2n]{\frac{1}{3^n}}\}$$
Then there are two subsequences of $(1)$ and $(2)$, each of which will lead to $\lim \sup$ or $\lim \inf$ for each sequence.
Questions:
1- Is my understanding right?
2- Why $\lim \sup \ (2)=\lim \sqrt[2n]{\frac{1}{2^n}}$ while $\{\sqrt[2n]{\frac{1}{2^n}}\}$ is not a subsequence of $(2)$?
3- Does $(1)$ or $(2)$ has only two subsequences?
4- Does $(1)$ or $(2)$ has only two subsequential limits?
5- Is there an informal way to choose the subsequence whose limit is $\lim \sup/\inf$ of a sequence, or should I consider every possible subsequence?
Your understanding is right.
The author just decided to write the limit in a way he likes, but it's not coming directly from the sequence; personally, I think it makes no sense to write it like that. To calculate the limit of the subsequence, $$ (a_{2n-1})^{1/(2n-1)}=(2^{-n})^{1/(2n-1)}=e^{-\frac{n}{2n-1}\,\log 2}\to e^{-\frac12\,\log 2}=2^{-1/2}=\frac1{\sqrt2}. $$
There are only two subsequential limits in this case, because the distance between $2^{-n/(2n-1)}$ and $3^{-1/2}$ cannot be small for large $n$.
There is no canonical recipe in general. These example works nicely because there are only two subsequences to choose from.