Why can we say If $\limsup \left | a_{n_{k}} \right |^{\frac{1}{n_{k}}}\geq 1\Rightarrow \limsup \left | a_{n} \right |^{\frac{1}{n}}\geq 1$ I have some questions about some specific parts of a solution that I found.
Problem: Consider a power series $\sum a_{n}x^{n}$ with radius of convergence $R$.  Prove that if $\limsup \left | a_{n} \right |>0$, then $R \leq 1$.
The solution that I found:
Since $\limsup \left | a_{n} \right |>0$, $\exists c$ s.t $\limsup \left | a_{n} \right |>c>0$.
Then $\forall N>0$, $\sup \left \{ |a_{n}|:n>N \right \}>c$.
Let $\left ( a_{n_{k}} \right )$ be a subsequence of $\left ( a_{n} \right )$.
Then, $\left | a_{n_{k}} \right |>c$ $\forall k$. $\cdot \cdot \cdot \bigstar $
$\Rightarrow \left | a_{n_{k}} \right |^{\frac{1}{n_{k}}}>c^{\frac{1}{n_{k}}}$.
By limit theorem, $\lim c^{\frac{1}{n_{k}}}=1$, so $\limsup  c^{\frac{1}{n_{k}}}=1$
$\Rightarrow \limsup \left | a_{n_{k}} \right |^{\frac{1}{n_{k}}}\geq \limsup c^{\frac{1}{n_{k}}}=1$ $\cdot \cdot \cdot \bigstar $
$\Rightarrow \limsup \left | a_{n} \right |^{\frac{1}{n}}\geq 1$ $\cdot \cdot \cdot \bigstar $
$\Rightarrow \beta \geq 1 \Rightarrow \frac{1}{\beta }=R\leq 1$ [End of Proof.]
I marked $\bigstar $ to indicate which part I have questions for.
First $\bigstar $: Why?
Second $\bigstar $: Can we take "$\limsup $" as we take "$\lim $" on inequalities?
Third  $\bigstar $: I think this part is the hardest part to understand since we usually get the property of a subsequence from a given property of a sequence. But, this is the opposite. It looks like it's saying "If lim sup of subsequence never goes below 1, then the lim sup of sequence also doesn't go below 1." Is this a general thing? Or is it a result of previous settings in the proof?
 A: Ok. I think I have the answer to most of these.

*

*I think that part of the confusion here is that it is written as if we are taking an arbitrary subsequence. But for the proof it is enough for there to be such a subsequence. The reason this is enough will be explained in the response to 4. Are you ok with the fact that we can find some subsequence $(a_{n_k})$ making this hold?

*We can apply $\limsup$ to inequalities just like we can $\lim$. This is because if we have $a_n > b_n$ for arbitrary sequences, then for any $m$, $\sup \{a_n: n > m\} \geq \sup \{b_n: n > m\}$ and so $\limsup a_n = \lim_{m\to \infty}\sup \{a_n: n > m\} \geq \lim_{m\to \infty}\sup \{b_n: n > m\} = \limsup b_n$.

*Generally, if you have some subsequence $b_{n_k}$ of a sequence $b_n$ then $\limsup b_n \geq \limsup b_{n_k}$. If you want more explanation I can talk more about this but it really is worth working out on your own. The reasoning has a similar flavor to a lot of what was done above. More generally, there are statements about the relationship between limits of subsequences of a sequence and the limit of a sequence. For example, it is true that if you have a sequence where every subsequence has a convergent sub-subsequence, and all of those convergent sub-subsequences have the same limit, then the sequence has that limit, as explained here.

