Power Series Converges Uniformly within Radius of Convergence

From ProofWiki
Jump to navigation Jump to search


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 $D = \set {x: \size {x - \xi} \le \rho}$.


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 $D$.