0
$\begingroup$

So my question is about the following Proposition: Let $\sum_n a_n (z-z_0)^n$ be a power series with radius of convergence $R$. Let $a_n \neq 0$ for all n. Then the following is true

$$\liminf |\frac{a_n}{a_{n+1}}| \leq R\leq \limsup |\frac{a_n}{a_{n+1}}|$$

Proof: define $S:=\liminf |\frac{a_n}{a_{n+1}}|$, let $0<s<S$ by the definition of lim inf there exists a $n_0$ such that $|a_n a^{-1}_{n+1}|>s$ for all $n>n_0$.

Now my Problem:

The Definition I know for $\liminf$ is as follows: let $a_n$ be a (bounded) sequence, the number $\alpha$ is called the limes inferior iff for every $\epsilon>0$ the inequality $$a_n<\alpha +\epsilon$$ holds for infinitely many $n$, and the inequality

$$a_n<\alpha - \epsilon$$ holds for at most for finite $n$.

If I use the definition I know I get: $|a_n a^{-1}_{n+1}| < S+\epsilon$ for infinitely many $n$, (*) and $$|a_n a^{-1}_{n+1}| < S- \epsilon < S$$ for finite $n$ .

Since the $s$ I choose is smaller $S$, how do I get to $|a_n a^{-1}_{n+1}|>s$ ?

$\endgroup$

1 Answer 1

1
$\begingroup$

Since $s<S$, $s=S-\varepsilon$, for some $\varepsilon>0$. So, the inequality$$\left|\frac{a_n}{a_{n+1}}\right|<s\tag1$$only holds for finitely many $n$'s. Let $N\in\Bbb N$ such that all those $n$'s for which $(1)$ holds are smaller than $N$. Then$$n\geqslant N\implies\left|\frac{a_n}{a_{n+1}}\right|\geqslant s.$$

$\endgroup$
2
  • $\begingroup$ Thank you very much! $\endgroup$
    – Andres2003
    Apr 3, 2022 at 21:08
  • 1
    $\begingroup$ I'm glad I could help. $\endgroup$ Apr 3, 2022 at 21:08

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .