Show that $\sum_{n = 0}^{\infty} x^{n} = \frac{1}{1 - x}$ where $|x| < 1$ is not uniformly convergent
My professor's proof is as follows:
So we know that the radius of convergence is $R = 1$. Now assume that $\sum_{n = 0}^{\infty} x^n$ is uniformly convergent on $(-1, 1)$, then $\forall 0 < \epsilon < 1, \exists N$, such that $n \geq N$ gives $|\sum_{k = n + 1}^{\infty} x^k| \leq \epsilon, \forall x \in (-1, 1)$ (*). So in particular, $x ^{n + 1} \leq \epsilon, \forall x \in (0, 1)$. Let $x \rightarrow 1$, then $1 \leq \epsilon$. Contradiction, so the series is not unif. convergent on $(-1, 1)$.
I got confused at (*), so to show that $f_{n}(x) = \sum_{n = 0}^{\infty} x^{n}$ is uniformly convergent, I must show that $\forall \epsilon > 0, \exists N$ for $n \geq N$ such that $|f_{n}(x) - f(x)| < \epsilon$. Since the professor wrote $|\sum_{k = n + 1}^{\infty} x^k| \leq \epsilon$, does that mean $f(x) = 0$ here? How so? Is it because $x^{n} \rightarrow 0$ for $|x| < 1$? So $\sum x^{n} \rightarrow 0$ as well?
Also, what's the point of noting that $\sum_{n = 0}^{\infty} x^{n} = \frac{1}{1 - x}$? I don't see $\frac{1}{1 - x}$ anywhere in the proof.
