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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?

john
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    Are you sure that $\forall N > 0, \exists n$ such that $\sup {a_n : n > N}$... is copied correctly? Because as it stands now, we are simply saying that we can find some $n>N$ such that $\sup {a_n} = a_n > c$. Or at least that's how I read it, and I'm almost certain this isn't what is wanted. – memerson Apr 08 '21 at 07:04
  • @memerson I just edited it to another version. There are two answers and I copied one of them. The edited one is the other one. But I still doubt why we are using such notation instead of just $\sup |a_{n}|$. – john Apr 08 '21 at 07:09
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    Hint for the question in title: If $A\subset B$, then $\limsup A\le \limsup B$. – Koro Apr 08 '21 at 07:11
  • @memerson After I changed it to the other version, it became obvious. So, I deleted a star next to it. – john Apr 08 '21 at 07:30

1 Answers1

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Ok. I think I have the answer to most of these.

  1. 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?
  2. 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 $\limsup a_n = \lim_{m\to \infty}\sup \{a_n: n > m\} \geq \lim_{m\to \infty}\sup \{b_n: n > m\} = \limsup b_n$.
  3. 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.
memerson
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  • Yes, it makes sense that we could always find some $|a_{n}| > c$, so there must exist some subsequence $(a_{n_{k}})$ of $(a_{n})$. Am I understanding it right? – john Apr 08 '21 at 07:47
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    @john, yeah. That's basically the gist of it – memerson Apr 08 '21 at 07:50
  • Then I also want to verify my understanding of the last question part. So, are we considering $(|a_{n}|^{\frac{1}{n}})$ and $(|a_{n_{k}}|^{\frac{1}{n_{k}}})$ as sequence and a subsequence here? Thus, $\limsup |a_{n}|^{\frac{1}{n}} \geq \limsup |a_{n_{k}}|^{\frac{1}{n_{k}}}$. Is this right? – john Apr 08 '21 at 08:02
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    You are correct. – memerson Apr 08 '21 at 08:07
  • Last follow up question is: we say $\limsup \left | a_{n} \right |^{\frac{1}{n}}\geq \limsup \left | a_{n_{k}} \right |^{\frac{1}{n_{k}}}$ simply because we have less term in $\left ( \left | a_{n_{k}} \right |^{\frac{1}{n_{k}}} \right )$. Is this also correct? – john Apr 08 '21 at 08:39
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    That's correct. – memerson Apr 08 '21 at 09:13