Show that \(\frac{1}{z}\sum_{k=1}^\infty \frac{z^k}{k}\) converges for all \(z \in S^1 \setminus \{1\}\) using summation by parts.
Solution
Fix \(z\in S^1\setminus\{1\}\) and set \[ A_N=\sum_{k=1}^N z^k. \] Since \[ A_N=z\frac{1-z^N}{1-z}, \] we have the uniform-in-\(N\) bound \[ |A_N|\le \frac{2}{|1-z|}. \]
Summation by parts gives \[ \sum_{k=1}^N\frac{z^k}{k} =\frac{A_N}{N} +\sum_{k=1}^{N-1}A_k\left(\frac1k-\frac1{k+1}\right). \] The first term tends to \(0\), while \[ \sum_{k=1}^\infty |A_k|\left(\frac1k-\frac1{k+1}\right) \le \frac{2}{|1-z|} \sum_{k=1}^\infty\frac1{k(k+1)}<\infty. \] Thus \(\sum z^k/k\) converges. Since \(z\ne0\), multiplication by \(1/z\) preserves convergence.