Interval of Convergence

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Theorem

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

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

Then the interval of convergence of $S \left({x}\right)$ is a real interval whose midpoint is $\xi$.

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.

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

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

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

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

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

The result follows.

$\blacksquare$

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

$\blacksquare$