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 (Dirichlet’s test).
Solution
Fix \(z\in S^1\setminus\{1\}\) and set \[ A_n=\sum_{k=1}^n z^k =z\frac{1-z^n}{1-z}. \] Since \(|z|=1\), \[ |A_n|\le\frac2{|1-z|} \] for every \(n\).
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 \[ \sum_{k=1}^\infty \left|A_k\left(\frac1k-\frac1{k+1}\right)\right| \le \frac2{|1-z|}\sum_{k=1}^\infty\frac1{k(k+1)}<\infty. \] Hence \(\sum_{k=1}^\infty z^k/k\) converges. Since \(z\ne0\) on \(S^1\), multiplication by \(1/z\) preserves convergence.