How to deal with lim sup and lim inf?

My analysis teacher taught us about these concepts using the ideas of "eventually" and "frequently". A sequence is "eventually" in a set if there exists some $N$ such that for every $n>N$, we have $a_n$ in the set. A sequence is "frequently" in a set if, for every $N$, there exists an $n>N$ with $a_n$ in the set.

Step one is to meditate on that paragraph until it makes sense with your understanding of the words "eventually" and "frequently".

Once you have that, $\lim\sup$ works this way. To say $\lim\sup\{a_n\}=c$ means that the sequence $\{a_n\}$ is eventually less than $c+\epsilon$, and frequently greater than $c-\epsilon$, for any small $\epsilon>0$.

The first claim, that the sequence is eventually less than $c+\epsilon$, means that there is no accumulation point of the sequence strictly larger than $c$. Adding in the other fact, that it is frequently greater than $c-\epsilon$, we know that $c$ is an accumulation point itself.

For $\lim\inf$, just turn all that on its head. If $\lim\inf\{a_n\}=c$, then $\{a_n\}$ is eventually greater than $c-\epsilon$, and it is frequently less than $c+\epsilon$, again for any small $\epsilon>0$.

Once you have these definitions in your head, you can gain intuition by playing around with practice sequences that have various kinds of limit points. You should be able to find such exercises in any decent analysis textbook. If you google "lim sup lim inf practice problems", you can find many more.

A nice thing about these concepts, "eventually" and "frequently", is that they can be easily adapted to talk about other kinds of limits in mathematics, such as limits of sequences of sets, and possibly even direct limits of algebraic and topological structures, although I have less experience with those, so I'm not too sure.

The number $\limsup|a_n|$ is the supremum of the accumulation points of the sequence $\{|a_n|\}$. So if you are told that $\limsup|a_n|=c>0$, you know that there is a subsequence $\{|a_{n_k}|\}$ with limit $c$, and no subsequence has a greater limit. The first fact is already enough to tackle your problem.

I would write here $\limsup|a_n|=k$. Then $k>0$. What this means is that for any $\delta>0$, $|a_n|>k-\delta$ infinitely often, but $|a_n|<k+\delta$ for all sufficiently large $n$. Thus $|a_n|^{1/n}>(k-\delta)^{1/n}$ infinitely often but $|a_n|<(k+\delta)^{1/n}$ for all sufficiently large $n$. Can we choose a useful $\delta$ which will tell us something about the sequence $(|a_n|^{1/n})$?

An idea for you to work out: that $\;\limsup\limits_{n\to\infty}a_n=q>0\;$ means that $\;q\;$ is the biggest of all the partial limits of $\;\{a_n\}\;$ , and in particular that there exists a subsequence

$$\{a_{n_k}\}\subset\{a_n\}\;\;\;s.t.\;\;\;\lim_{x\to\infty}a_{n_k}=q$$

and this last means that for $\;\epsilon=\frac q2\;$ , for example, there exists $\;K\in\Bbb N\;$ such that if $\;k>K\;$ then

$$\;|a_{n_k}-q|<\epsilon\iff 0<q-\epsilon<a_{n_k}<q+\epsilon$$

and from here

$$1\xleftarrow[\infty\leftarrow n]{}\sqrt[n]{q-\epsilon}\le\sqrt[n]{a_{n_k}}\le\sqrt[n]{q+\epsilon}\xrightarrow[n\to\infty]{}1$$

Deduce now what you asked...