Use summation by parts to show that \(\sin(n)/n\) converges.
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
Use that \(\abs{z^k} = 1\) and \(\sum c_kz^k < \infty \implies \abs{c_k} \to 0\), but \(\abs{kz^k} = \abs{k} \to \infty\) here.
Use that absolutely convergent implies convergent, and \(\sum \abs{k^{-2} z^k} = \sum \abs{k^{-2}}\) converges by the \(p\dash\)test.
If \(z=1\), this is the harmonic series. Otherwise take \(a_k = 1/k, b_k = e^{i k \theta}\) where \(\theta \in (0, 2\pi)\) is some constant, and apply Dirichlet’s test. It suffices to bound the partial sums of the \(b_k\). Recalling that \(\sum_{k\leq N} r^k = (1-r^{N+1}) / (1-r)\), \[ \norm{ \sum_{k\leq m} e^{ik\theta } } = \norm{1 - e^{i(m+1)\theta} \over 1 - e^{i\theta}} \leq {2 \over \norm{ 1- e^{i\theta}}} \da M ,\] which is a constant. Here we’ve used that two points on \(S^1\) are at most distance 2 from each other.