Existence of Interval of Convergence of Power Series

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

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

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

Hence, from Convergent Sequence is Bounded, $$\left \langle {a_n \left({y - \xi}\right)^n} \right \rangle$$ is bounded.

Thus $$\exists H \in \R: \forall n \in \N^*: \left|{a_n \left({y - \xi}\right)^n}\right| \le H$$.

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

Then $$\rho = \frac{\left|{x - \xi}\right|} {\left|{y - \xi}\right|} < 1$$.

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

Hence $$\forall n \in \N^*: \left|{a_n \left({x - \xi}\right)^n}\right| = \rho^n \left|{a_n \left({y - \xi}\right)^n}\right| \le H \rho^n$$.

By Power of a Number Less Than One, $$\sum_{n=1}^\infty \rho^n$$ converges.

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

The result follows.

Proof of Corollary
Follows directly from the method of proof of the main result.