So I'm looking at the proof of the Cauchy-Hadamard formula in my textbook and I understand the first part, but the second one I'm confused about. Here goes: $$\DeclareMathOperator*\uplim{\overline{lim}} \text{Let } \sum_{n=0}^{\infty}a_n(x-x_0)^n \text{ be a power series, and let } \rho=\uplim_{n \to \infty}{\sqrt[n]{|a_n|}} \text{. Then}$$ (1) if $\rho \in(0, \infty)$ $R = \frac{1}{\rho}$
(2) if $\rho = +\infty$ $R=0$
(3) if $\rho=0$ $R = +\infty $
The proofs for (2) and (3) are beyond the scope of our course, so we only did proof for (1). This is the first part:
Let $t\in \mathbb{R}$, and suppose $|t-x_0|>\frac{1}{\rho}$. Then $\frac{1}{|t-x_o|}<\rho$, therefore we can say $(\exists\epsilon>0)$ such that $\frac{1}{|t-x_0|} < \rho-\epsilon \Rightarrow (\rho-\epsilon)|t-x_0|>1$.
Since $\rho=\uplim_{n \to \infty}{\sqrt[n]{|a_n|}}$ then $\sqrt[n]{|a_n|}>\rho-\epsilon$ for all $n\geq n_o$, $n_0 \in \mathbb{N}$. Therefore:
$$|a_n(t-x_0)^n| = \bigg|\sqrt[n]{|a_n|}(t-x_0)\bigg|^n>((\rho-\epsilon)|t-x_0|)^n > 1$$
So we have that $|a_n(t-x_0)^n| > 1$ meaning that it cannot converge to zero, meaning that the series diverges. So we conclude that:
$$|t-x_0|>\frac{1}{\rho} \Rightarrow \text{the series diverges}$$
Taking the counterposition of this gives us $$\text{the series converges} \Rightarrow |t-x_0|\leq\frac{1}{\rho}\Rightarrow R\leq\frac{1}{\rho}$$
And all this is clear so far. Now we have to show that also $R\geq\frac{1}{\rho}$ which will get us to the final conclusion that $R = \frac{1}{\rho}$
Let's suppose otherwise $R < \frac{1}{\rho}$
Now here's the problematic part (I'm quoting now):
then we know $(\exists t\in \mathbb{R})$ such that $R<|t-x_0|<\frac{1}{\rho}$. I have no idea how this was concluded. After this we have something similar to the first part:
$$\frac{1}{|t-x_0|}>\rho \Rightarrow (\exists \epsilon>0) \quad \frac{1}{|t-x_0|} >\rho + \epsilon \Rightarrow (\rho+\epsilon)|t-x_0|<1$$
Because of how we defined $\rho$ we got:
$$(\exists n_0 \in \mathbb{N})(\forall n \geq n_0)(\sqrt[n]{|a_n|}<\rho +\epsilon)$$ Then:
$$\sqrt[n]{|a_n|(t-x_0)^n} = \sqrt[n]{|a_n|}|t-x_0| < (\rho + \epsilon)|t-x_0| = q$$ where $|q|<1$.
Now we have that $a_n(t-x_0)^n < q^n$ where $|q|<1$ meaning $q$ converges, meaning $a_n(t-x_0)^n$ converges (by the comparison criteria). Since $R$ is defined as $$R = \sup{\bigg\{|t-x_0| \text{ the sequence converges for }t\bigg\}}$$
And since our sequence indeed converges for $t$, we have (the above mentioned set we'll call $E$). $$|t-x_0| \in E \Rightarrow |t-x_0| \leq \sup{E} = R $$
which gives us a contradiction with one of the beginning statements (i.e., the one above whose justification I don't understand). So could anyone clarify? Thanks.
Let $R, \rho, x_0$ be any reals such that $R< \frac{1}{\rho}$. Pick any $a \in \mathbb{R}$ such that $R<a<\frac{1}{\rho}$ and set $|t-x_0| = a$ (for example $t=a+x_0$). Then there exists a $t>0$ such that $R < |t-x_0| < \frac{1}{\rho}$.
This is just the density of the real numbers: between any two reals there is an interval of real numbers.