About $\max |\sum_{n=0}^{n=\infty} a_n x^n - \sum_{n=0}^{n=N} a_n x^n| \to 0$ ($N \to \infty$) I am reading "A First Course in Analysis vol.1" by Sin Hitotumatu.
The author writes the following corollary:

Corollary:
A power series with a positive radius of convergence is continuous in the circle of convergence.

Then he writes the following fact without a proof as a remark.

Let $\rho$ be the radius of the circle of convergence of $\sum_{n=0}^{n=\infty} a_n x^n$.
Let $f(x) := \sum_{n=0}^{n=\infty} a_n x^n$.
Let $f_N(x) := \sum_{n=0}^{n=N} a_n x^n$.
Let $\alpha$ be a real number such that $0 < \alpha < \rho$.
Then $\max_{x \in [-\alpha, \alpha]} |f(x) - f_N(x)| \to 0$ ($N \to \infty$).

I guess this fact follows easily from the above corollary.
Please tell me the proof.
 A: The corollary follows from the fact -- which is a well-known theorem for power series.
The "fact" tells you that the power series converges uniformly on the compact interval $[-\alpha, \alpha]$ inside the interval of convergence $(-\rho,\rho)$.
The concluding statement about the limit implies that for any $\epsilon> 0$ there exists $M(\epsilon) \in \mathbb{N}$ such that for all $N > M(\epsilon)$ and for all $x \in [-\alpha,\alpha]$, we have
$$|f(x) - f_N(x)| \leqslant \max_{x \in [-\alpha,\alpha]}|f_n(x) - f(x)| < \epsilon$$
To proceed, either invoke a well-known theorem that directly states that $f$ must be continuous at any point $c \in [-\alpha, \alpha]$ or prove it directly as follows.  
Taking $x,c \in [-\alpha,\alpha]$, we have 
$$\tag{*}|f(x) - f(c) | \leqslant \left|f(x) - \sum_{k=0}^na_kx^k\right|+ \left| \sum_{k=0}^na_kx^k- \sum_{k=0}^na_kc^k \right|+\left|f(c) - \sum_{k=0}^na_kc^k\right|$$
By uniform convergence of the power series for any $\epsilon > 0$ there exists $n > M(\epsilon/3)$ such that for any $x,c \in [-\alpha, \alpha]$ the first and third terms on the RHS of (*) are each leass than $\epsilon/3$.
Hence,
$$|f(x) - f(c) | \leqslant \frac{2\epsilon}{3} + \underbrace{\left| \sum_{k=0}^n a_kx^k- \sum_{k=0}^na_kc^k \right|}_{|P_n(x) - P_n(c)|}$$
With $n$ fixed the polynomial $P_n(x)= \sum_{k=0}^na_kx^k$ is continuous everywhere, and at $c$, in particular. Hence, there exists $\delta(\epsilon,c) > 0$ such that if $|x-c| < \delta(\epsilon,c)$ then $|P_n(x) - P_n(c)|< \epsilon/3$
Thus, if $|x-c| < \delta(\epsilon,c)$, then $|f(x) - f(c)| < \epsilon$ and $f$ is continuous at any $c \in [-\alpha,\alpha]$.
