# Existence of Interval of Convergence of Power Series

## Theorem

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

Let $\displaystyle \map S x = \sum_{n \mathop = 0}^\infty a_n \paren {x - \xi}^n$ 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 $\displaystyle S \left({x}\right) = \sum_{n \mathop = 0}^\infty a_n x^n$ 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 $\map S x$ 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 $x = y$.

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

Hence, from Convergent Sequence is Bounded:

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

Thus:

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

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

Then:

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

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

Hence:

$\forall n \in \N_{>0} \size {a_n \paren {x - \xi}^n} = \rho^n \size {a_n \paren {y - \xi}^n} \le H \rho^n$
$\displaystyle \sum_{n \mathop = 1}^\infty \rho^n$ converges

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

The result follows.

$\blacksquare$