Suppose \(\sum_{n=0}^\infty a_n z^n\) converges for some \(z_0 \neq 0\).
Prove that the series converges absolutely for each \(z\) with \(\abs z < \abs z_0\).
Suppose \(0 < r < \abs{z_0}\) and show that the series converges uniformly on \(\abs{z} \leq r\).
Solution
Part (a).
1. Since \(\sum a_n z_0^n\) converges, \(a_n z_0^n \to 0\), so \(\{a_n z_0^n\}\) is bounded: \(|a_n z_0^n| \le M\) for some \(M\).
the terms of a convergent series tend to \(0\), hence are bounded.
2. For \(|z| < |z_0|\), \(|a_n z^n| = |a_n z_0^n| \left|\frac{z}{z_0}\right|^n \le M \left|\frac{z}{z_0}\right|^n\).
1.
3. \(\sum M \left|\frac{z}{z_0}\right|^n\) converges (geometric series with ratio \(|z/z_0| < 1\)).
geometric series.
4. Hence \(\sum |a_n z^n|\) converges by the comparison test.
2 and 3.
Part (b).
1. For \(|z| \le r < |z_0|\), \(|a_n z^n| \le M \left(\frac{r}{|z_0|}\right)^n\).
2 of part (a), with \(|z| \le r\).
2. \(\sum M \left(\frac{r}{|z_0|}\right)^n\) converges (geometric series with ratio \(r/|z_0| < 1\)).
geometric series.
3. Hence \(\sum a_n z^n\) converges uniformly on \(|z| \le r\) by the Weierstrass \(M\)-test.
1 and 2.
4. Q.E.D.
4 (a) and 3 (b).