# Question about the proof of quotient ratio test

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 $$0 by the definition of lim inf there exists a $$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 $$|a_n a^{-1}_{n+1}|>s$$ ?

Since $$s, $$s=S-\varepsilon$$, for some $$\varepsilon>0$$. So, the inequality$$\left|\frac{a_n}{a_{n+1}}\right|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$$n\geqslant N\implies\left|\frac{a_n}{a_{n+1}}\right|\geqslant s.$$