- Show that any power series is continuous on its domain of convergence.
Solution
Goal: Show that a power series \(\sum a_n z^n\) is continuous on its domain of convergence (the disk \(\abs z < R\), where \(R\) is its radius of convergence).
1. Fix \(z_0\) with \(\abs{z_0} < R\), and choose \(\rho\) with \(\abs{z_0} < \rho < R\).
The interval \((\abs{z_0}, R)\) is nonempty, so such \(\rho\) exists.
2. The series converges absolutely and uniformly on the closed disk \(\abs z \leq \rho\).
2.1. \(\sum_{n} \abs{a_n} \rho^n < \infty\).
A power series converges absolutely inside its radius of convergence; since \(\rho < R\), the series at \(\rho\) converges absolutely.
2.2. \(\abs{a_n z^n} \leq \abs{a_n} \rho^n\) for all \(\abs z \leq \rho\).
\(\abs z^n \leq \rho^n\).
2.3. Weierstrass \(M\)-test.
1 and 2 give uniform convergence on \(\abs z \leq \rho\).
3. The partial sums \(S_N(z) := \sum_{n=0}^{N} a_n z^n\) are continuous on \(\abs z \leq \rho\).
Polynomials are continuous.
4. The uniform limit of continuous functions is continuous.
Standard epsilon/3 argument using uniform convergence (<1>2.3).
5. Hence the power series is continuous at \(z_0\).
\(z_0\) lies in the interior of \(\abs z \leq \rho\) (<1>1), and the series is continuous on that closed disk (<1>2–<1>4).
6. Q.E.D.
\(z_0\) was arbitrary with \(\abs{z_0} < R\), so the series is continuous on its whole domain of convergence.