Existence of Interval of Convergence of Power Series

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Theorem

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

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


Then the interval of convergence of $S \left({x}\right)$ 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 $\displaystyle S \left({x}\right) = \sum_{n \mathop = 0}^\infty a_n x^n$ be a power series.

Let $S \left({x}\right)$ be convergent at $x = x_0$.


Then $S \left({x}\right)$ is convergent for all $x$ such that $\left\lvert{x}\right\rvert < \left\lvert{x_0}\right\rvert$.


Proof

Suppose $S \left({x}\right)$ converges when $x = y$.

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


So, let $S \left({x}\right)$ converge when $x = y$.

Then from Terms in Convergent Series Converge to Zero:

$a_n \left({y - \xi}\right)^n \to 0$ as $n \to \infty$

Hence, from Convergent Sequence is Bounded:

$\left \langle {a_n \left({y - \xi}\right)^n} \right \rangle$ is bounded

Thus:

$\exists H \in \R: \forall n \in \N_{>0}: \left|{a_n \left({y - \xi}\right)^n}\right| \le H$


Now suppose $\left|{x - \xi}\right| < \left|{y - \xi}\right|$.

Then:

$\rho = \dfrac{\left|{x - \xi}\right|} {\left|{y - \xi}\right|} < 1$

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

Hence:

$\forall n \in \N_{>0} \left|{a_n \left({x - \xi}\right)^n}\right| = \rho^n \left|{a_n \left({y - \xi}\right)^n}\right| \le H \rho^n$

By Sequence of Powers of Number less than One:

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

Thus $\displaystyle \sum_{n \mathop = 0}^\infty a_n \left({x - \xi}\right)^n$ converges by the Comparison Test.

The result follows.

$\blacksquare$


Also see


Sources