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.
Both tests are for series with positive terms; or you should put absolute values around $A_n$ and $A_{n+1}$ everywhere. Otherwise you'll have a problem taking the roots. And division by $0$ is generally frowned upon.
Assuming all the terms are positive, it is true that
$$\liminf\limits_{n\rightarrow \infty} \frac{A_{n+1}}{A_n}
\leq
\liminf\limits_{n\rightarrow \infty} (A_n)^{1/n}
\leq
\limsup\limits_{n\rightarrow \infty} (A_n)^{1/n} \leq \limsup_{n\rightarrow \infty} \frac{A_{n+1}}{A_n}$$
in the extended sense: the limits are allowed to take value $+\infty$. (By the way, this is common with $\liminf $ and $\limsup$ anyway.
The proof is not really different from the finite case. Here's a proof of the first inequality. Let $b$ be a number such that $b< \liminf\limits_{n\rightarrow \infty} \frac{A_{n+1}}{A_n} $. Then we have $A_{n+1}>bA_n$ for all sufficiently large $n$. Applying this iteratively, you'll get a lower bound of the form $A_n\ge C b^n$ for all sufficiently large $n$. This implies $\liminf\limits_{n\rightarrow \infty} (A_n)^{1/n} \ge b$. Since $b$ was an arbitrary number such that $b< \liminf\limits_{n\rightarrow \infty} \frac{A_{n+1}}{A_n} $, the conclusion $\liminf\limits_{n\rightarrow \infty} (A_n)^{1/n}\ge \liminf\limits_{n\rightarrow \infty} \frac{A_{n+1}}{A_n}$ follows.
Note that it makes no difference whether these limits are finite or not; the argument works the same.
Best Answer
If $\lim$ exists, it is of course the same as $\limsup$. The formula with $\limsup$ is useful when $\lim$ doesn't exist. Example: With $a_n=\frac{1+(-1)^n}2$ the expression$\sqrt[n]{|a_n|}$ has no limit, but the $\limsup$ is $1$.