Consider a (complex) power series \(P(z)=\sum_{n=0}^{\infty} a_n(z-z_0)^n\). Assume there are numbers \(z_1\in \mathbb{C}\) and \(M>0\) such that \(\lvert a_nz^n\rvert\le M\) for all \(n\in \mathbb{N}\). Then \(P\) converges locally absolutely uniformly on the disk \(D_{\lvert z_1\rvert}(z_0)\).

Proof

Let \(z_2\in \mathbb{C}\) such that \(\lvert z_2-z_0\rvert < \lvert z_1-z_0\rvert\). Then \(q:=\lvert \frac{z_2-z_0}{z_1-z_0}\rvert <1\) and

\begin{equation*} \lvert a_n (z_2-z_0)^n\rvert = \lvert a_n (z_1-z_0)^n\rvert q^n \le Mq^n. \end{equation*}

By comparison with the geometric series \(P(z_2)=\sum_{n=0}^{\infty} a_n (z_2-z_0)^n\) converges absolutely .

By modifying the above argument we obtain the local absolute uniform convergence .

Remarks
  • There is also a multidimensional version of Abel’s Lemma [1, Proposition 2.1.7]

References Link to heading

  1. S. Krantz and H. Parks, A Primer of Real Analytic Functions. Boston, MA: Birkhäuser Boston, 2002. doi:10.1007/978-0-8176-8134-0