Show that
\(\sum kz^k\) diverges on \(S^1\).
\(\sum k^{-2} z^k\) converges on \(S^1\).
\(\sum \inverseof{k} z^k\) converges on \(S^1\sm\ts{1}\) and diverges at \(1\).
Solution
1. (1) \(\sum kz^k\) diverges at every \(z\in S^1\).
The terms of a convergent series tend to \(0\), but \(\abs{kz^k} = k \to \infty\) for \(\abs z=1\).
2. (2) \(\sum k^{-2} z^k\) converges at every \(z\in S^1\).
For \(\abs z=1\), \(\sum \abs{k^{-2} z^k} = \sum k^{-2}\) converges by the \(p\)-test, and an absolutely convergent series converges.
3. (3) \(\sum k^{-1} z^k\) diverges at \(z=1\) and converges at every \(z\in S^1\setminus\{1\}\).
At \(z=1\) the series is the harmonic series. Otherwise \(z = e^{i\theta}\) with \(\theta \in (0, 2\pi)\). The partial sums of \(\sum e^{ik\theta}\) are bounded, since the geometric sum formula gives \[\abs{\sum_{k=0}^{m} e^{ik\theta}} = \abs{\frac{1 - e^{i(m+1)\theta}}{1 - e^{i\theta}}} \leq \frac{2}{\abs{1- e^{i\theta}}}\] for every \(m\), using \(\abs{1 - e^{i(m+1)\theta}}\le2\). The coefficients \(1/k\) decrease monotonically to \(0\), so Dirichlet’s test gives convergence.