Existence of Interval of Convergence of Power Series

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Let $\xi \in \R$ be a real number.

Let $\ds \map S x = \sum_{n \mathop = 0}^\infty a_n \paren {x - \xi}^n$ be a power series about $\xi$.

Then the interval of convergence of $\map S x$ is a real interval whose midpoint is $\xi$.

Corollary 1

A power series converges absolutely at all points of its interval of convergence with the possible exception of its end points.

At the end points nothing can be said: it could be absolutely convergent, or conditionally convergent, or divergent.

Corollary 2

Let $\ds \map S x = \sum_{n \mathop = 0}^\infty a_n x^n$ be a power series.

Let $\map S x$ be convergent at $x = x_0$.

Then $\map S x$ is convergent for all $x$ such that $\sequence x < \sequence {x_0}$.


Suppose $\map S x$ converges when $x = y$.

We need to show that it converges for all $x$ which satisfy $\size {x - \xi} < \size {y - \xi}$.

So, let $\map S x$ converge when $x = y$.

Then from Terms in Convergent Series Converge to Zero:

$a_n \paren {y - \xi}^n \to 0$ as $n \to \infty$

Hence, from Convergent Sequence is Bounded:

$\sequence {a_n \paren {y - \xi}^n}$ is bounded


$\exists H \in \R: \forall n \in \N_{>0}: \size {a_n \paren {y - \xi}^n} \le H$

Now suppose $\size {x - \xi} < \size {y - \xi}$.


$\rho = \dfrac {\size {x - \xi} } {\size {y - \xi} } < 1$

(Note that if $\size {x - \xi} < \size {y - \xi}$ then $\size {y - \xi} > 0$ and the above fraction always exists.)


$\forall n \in \N_{>0} \size {a_n \paren {x - \xi}^n} = \rho^n \size {a_n \paren {y - \xi}^n} \le H \rho^n$

By Sequence of Powers of Number less than One:

$\ds \sum_{n \mathop = 1}^\infty \rho^n$ converges

Thus $\ds \sum_{n \mathop = 0}^\infty a_n \paren {x - \xi}^n$ converges by the Comparison Test.

The result follows.


Also see