Rudin Series ratio and root test. In Rudins Principles of Mathematical Analysis he says consider the following series
$$\frac 12 + \frac 13 + \frac 1{2^2} + \frac 1{3^2} + \frac 1{2^3} + \frac 1{3^3} + \frac 1{2^4} + \frac 1{3^4} + \cdots$$
for which 
$$\liminf \limits_{n \to \infty} \dfrac{a_{n+1}}{a_n} =  \lim \limits_{n \to \infty} \left( \dfrac {2}{3} \right)^n =0, $$
$$\liminf \limits_{n \to \infty} \sqrt[n]{a_n} =  \lim \limits_{n \to \infty} \sqrt[2n]{\dfrac{1}{3^n}} = \dfrac{1}{\sqrt{3}}, $$
$$\limsup \limits_{n \to \infty} \sqrt[n]{a_n} =  \lim \limits_{n \to \infty} \sqrt[2n]{\dfrac{1}{2^n}} = \dfrac{1}{\sqrt{2}}, $$
$$\limsup \limits_{n \to \infty} \dfrac{a_{n+1}}{a_n} =  \lim \limits_{n \to \infty} \dfrac 12\left( \dfrac {3}{2} \right)^n =+\infty, $$
The root test indicates convergence; the ratio test does not apply.

In the book he defines the root and ratios test for the lim sup.  I am not exactly sure how he goes from the lim sup to the lim and also why there is a $2n$ (which I assume comes from even terms of the sequence) in the root test.  Also why is he checking the lim inf?  I believe that my understanding of lim sups and infs are not well developed or I would probably understand what’s going on.
Also how does he get the terms that he is taking the limit of.  A nudge in the right direction to figure this out would be much appreciated.     Thank you!!

For definiteness, call the terms of our sequence $a_1,a_2,a_3,\dots$. A similar analysis with minor differences of detail can be made if we call the first term of our sequence $a_0$.
Note that for $n=1,2,3,\dots$ we have $a_{2n-1}=\dfrac{1}{2^n}$ and $a_{2n}=\dfrac{1}{3^n}$. 
The $k$-th root of the $k$-th term is "small" when the $k$-th term is a power of $\dfrac{1}{3}$. The $k$-th root of the $k$-th term is "large" when the $k$-th term is a power of $\dfrac{1}{3}$.
More precisely, $\liminf \sqrt[k]{a_k}=\lim\inf \sqrt[2n]{\frac{1}{3^n}}=\dfrac{1}{3}$.  For even $k$ the $k$-th root is constant. 
Also, $\limsup\sqrt[k]{a_k}=\liminf\sqrt[2n-1]{\dfrac{1}{2^n}}$. But 
$$\sqrt[2n-1]{\dfrac{1}{2^n}}=\left(\frac{1}{2^n}\right)^{1/(2n-1)}=\left(\frac{1}{2^n}\right)^{2n/(2n(2n-1))}=\left(\frac{1}{\sqrt{2}}\right)^{2n/(2n-1)}.$$
The expression on the right has limit $\dfrac{1}{\sqrt{2}}$.
That takes care of one of the gaps.
For the Ratio Test, we are interested in the behaviour of $\left|\dfrac{a_{k+1}}{a_k}\right|$. 
Let $k$ be odd, say $k=2n-1$. Then $a_k=\dfrac{1}{2^n}$. And $a_{k+1}=a_{2n}=\dfrac{1}{3^n}$. It follows that
$$\frac{a_{k+1}}{a_k}=\frac{a_{2n}}{a_{2n-1}}=\left(\frac{2}{3}\right)^n.$$
This has very pleasant behaviour for large $n$, indeed for any $n$: it is safely under $1$, indeed has limit $0$.
Now let $k$ be even, say $k=2n$. Then $a_k=\dfrac{1}{2^n}$. and $k+1=2n+1$. The $2n+1$-th term of our sequence is $\dfrac{1}{2^{n+1}}$. It follows that in the case $k=2n$ we have
$$\frac{a_{k+1}}{a_k}=\frac{a_{2n+1}}{a_{2n}}=\frac{\frac{1}{2^{n+1}}}{\frac{1}{3^n}}=\frac{1}{2}\left(\frac{3}{2}\right)^n.$$
This unfortunately behaves badly for large $n$: we would like it to be safely under $1$, and it is very much over.
The limit of the ratios $\dfrac{a_{k+1}}{a_k}$ does not exist. The ratios do not (uniformly) blow up, since for $k$ odd, the ratios approach $0$. The ratio behaves very nicely at odd $k$, and very badly at even $k$. So the Ratio Test is inconclusive. The bad behaviour prevents us from concluding convergence. But the good behaviour prevents us from concluding divergence.
