Power Series Converges Uniformly within Radius of Convergence

Theorem
Let $\displaystyle S := \sum_{n \mathop = 0}^\infty a_n \left({x - \xi}\right)^n$ be a power series about a point $\xi$.

Let $R$ be the radius of convergence of $S$.

Let $\rho \in \R$ such that $0 \le \rho < R$.

Then $S$ is uniformly convergent on $D = \left\{{x: \left|{x - \xi}\right| \le \rho}\right\}$.

Proof
We shall make use of the Weierstrass M-Test to prove this result.

To begin with, for each $n \in N$, define $f_n(x) = a_n (x - \xi)^n$ where $x \in D$.

Since $x \in D$, it follows that $|x - \xi| \le \rho$ and, therefore, $|f_n(x)| = |a_n (x - \xi)^n| \le |a_n \rho^n|$ for every $x \in D$.

Defining $M_n = |a_n \rho^n|$, we, thus, see that $M_n$ is an upper bound on $|f_n(x)|$.

So, by the definition of a supremum:
 * $\displaystyle \sup_{x \mathop \in D} |f_n(x)| \le M_n$

Next, to apply the Weierstrass M-Test, it remains to be shown that $\displaystyle \sum_{n \mathop = 0}^\infty M_n = \sum_{n \mathop = 0}^\infty |a_n \rho^n|$ converges.

Well, $\displaystyle \sum_{n \mathop = 0}^\infty a_n \rho^n$ certainly converges.

To see this, just choose $x = \xi + \rho$ and note that $|x-\xi|=|\rho|=\rho < R$.

Hence, by Definition:Radius of Convergence, $x$ falls within the interval of convergence of $S$ and $\displaystyle S = \sum_{n \mathop = 0}^\infty a_n (x - \xi)^n = \sum_{n \mathop = 0}^\infty a_n \rho^n$ converges.

By Existence of Interval of Convergence of Power Series, a power series always converges absolutely at all points in its interval of convergence.

Hence $\displaystyle \sum_{n \mathop = 0}^\infty M_n = \sum_{n \mathop = 0}^\infty |a_n \rho^n|$ also converges.

Finally then, having satisfied all the requirements to use Weierstrass M-Test, we can conclude that $\displaystyle \sum_{n \mathop = 0}^\infty f_n(x) = \sum_{n \mathop = 0}^\infty a_n (x - \xi)^n$ converges uniformly in $D$.