📂Analysis

Convergence of Power Series

Theorem¹

Power Series $\sum\limits_{n = 0}^{\infty} c_{n} (x - a)^{n}$’s radius of convergence is $R$. Then,

1. The series converges absolutely within $x \in (a - R, a + R)$.
2. The series converges uniformly within any closed interval $[b, d] \subset (a - R, a + R)$.
3. Regarding $(R \lt \infty$, the series diverges beyond $x \notin [a - R, a + R]$.

Explanation

Refer to here for the proof of 1 and 3.

The statement 2 can be restated as follows:

For any positive number $\varepsilon \gt 0$, the series converges uniformly within $[a - R + \varepsilon, a + R - \varepsilon]$.

Proof (2.)

Suppose $\varepsilon \gt 0$ is given. For $\left| x - a \right| \le R - ε$,

$$\left| c_{n} (x - a)^{n} \right| \le \left| c_{n} (R - ε)^{n} \right|$$

holds true. However, by 1, the series $\sum\limits_{n = 0}^{\infty} c_{n} (R - ε)^{n}$ converges absolutely. If we let $M_{n} = \left| c_{n} (R - ε)^{n} \right|$, then by the Weierstrass-$M$ Test, $\sum\limits_{n = 0}^{\infty} c_{n} (x - a)^{n}$ converges uniformly within $[a - R + \varepsilon, a + R - \varepsilon]$.

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