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Power Series

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1 Definition

Let \( (a_n)_n \) be some sequence, then a series of the form:

\[ \sum_{n=1}^{\infty} a_n x^n \]

Is called a power series (in \( x \)), where \( x \in \mathbb{R} \).

We make the statement that one of the following must be true:

\[ S = \sum_{n=1}^{\infty} a_n x^n \]

Converges:

  1. \( \forall x \in \mathbb{R} \)
  2. Only for \( x = 0 \)
  3. \( \forall |x| \le R \text{, and diverges } \forall |x| > R \ where \( R \) is some real. We this \( R \) The Radius of Convergence of the power series

1.1 Proof

Suppose \( S \) converges for some \( x = r \), then we know that the sequence \( (a_n r^n)_n \) is null and hence that \( \exists M \ s.t. \ |a_n r^n| \le M \ \forall n \). Now consider some \( |u| < |r| \), and let \( t = \frac{|u|}{|r|} \). So that \( 0 < t < 1 \) and

\[ \mid a_n u^n \mid \ = \ \mid a_n r^n \mid t^n \le M t^n \ \forall n \]

Summing over infinity:

\[ \implies 0 \le \sum_{n=1}^{\infty} |a_n u^n| \le M\sum_{n=1}^{\infty} t^n \]

Then the series on the right is a convergent geometric series by the predicate on \( t \), and hence we can conclude by the sandwich rule that \( S \) converges absolutely for \( x = u \).

Author: root

Created: 2024-03-23 Sat 11:44

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