Existence of Interval of Convergence of Power Series

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 $\xi$.

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 Power 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.

Also see

 * Radius of Convergence of Complex Power Series for a proof of the same result in complex numbers.