Power Series Converges Uniformly within Radius of Convergence
Theorem
Let $\ds S := \sum_{n \mathop = 0}^\infty a_n \paren {x - \xi}^n$ 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 $D = \set {x: \size {x - \xi} \le \rho}$.
Proof
We shall make use of the Weierstrass M-Test to prove this result.
To begin with, for each $n \in N$, define for $x \in D$:
- $\map {f_n} x = a_n \paren {x - \xi}^n$
We have:
\(\ds \size {\map {f_n} x}\) | \(=\) | \(\ds \size {a_n \paren {x - \xi}^n}\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \size {a_n \rho^n}\) | $x \in D$ |
Define $M_n = \size {a_n \rho^n}$.
Then $M_n$ is an upper bound for $\size {\map {f_n} x}$.
So, by the definition of a supremum:
- $\ds \sup_{x \mathop \in D} \size {\map {f_n} x} \le M_n$
Next, to apply the Weierstrass M-Test, it remains to be shown that $\ds \sum_{n \mathop = 0}^\infty M_n = \sum_{n \mathop = 0}^\infty \size {a_n \rho^n}$ converges.
Choose $x = \xi + \rho$ and note that:
- $\size {x - \xi} = \size \rho = \rho < R$
Hence, by definition of radius of convergence, $x$ falls within the interval of convergence of $S$.
Thus $\ds S = \sum_{n \mathop = 0}^\infty a_n (x - \xi)^n = \sum_{n \mathop = 0}^\infty a_n \rho^n$ 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 $\ds \sum_{n \mathop = 0}^\infty M_n = \sum_{n \mathop = 0}^\infty \size {a_n \rho^n}$ also converges.
Finally then, having satisfied all the requirements to use Weierstrass M-Test, we can conclude that $\ds \sum_{n \mathop = 0}^\infty \map {f_n} x = \sum_{n \mathop = 0}^\infty a_n \paren {x - \xi}^n$ converges uniformly in $D$.
$\blacksquare$
Sources
- 1992: Larry C. Andrews: Special Functions of Mathematics for Engineers (2nd ed.) ... (previous) ... (next): $\S 1.3.2$: Power series