Prove the following:
\(\sum_{n} nz^n\) does not converge at any point of \(S^1\)
\(\sum_n {z^n \over n^2}\) converges at every point of \(S^1\).
\(\sum_n {z^n \over n}\) converges at every point of \(S^1\) except \(z=1\).
Solution
Goal: Prove: (a) \(\sum_{n} nz^n\) does not converge at any point of \(S^1\); (b) \(\sum_n z^n/n^2\) converges at every point of \(S^1\); (c) \(\sum_n z^n/n\) converges at every point of \(S^1\) except \(z = 1\).
1. Part (a): for \(\abs z = 1\), the terms do not tend to \(0\).
\(\abs{n z^n} = n \cdot \abs z^n = n \to \infty\) as \(n \to \infty\); a necessary condition for convergence of a series is that its terms tend to \(0\).
2. Part (b): \(\sum_n \frac{z^n}{n^2}\) converges absolutely at every point of \(S^1\).
2.1. \(\abs{\frac{z^n}{n^2}} = \frac{1}{n^2}\) for \(\abs z = 1\).
\(\abs{z^n} = 1\).
2.2. \(\sum_n \frac{1}{n^2}\) converges.
\(p\)-series with \(p = 2 > 1\).
2.3. Hence the series converges absolutely, in particular converges, at every \(z \in S^1\).
Absolute convergence (comparison test, <2>1–<2>2) implies convergence.
3. Part (c) at \(z = 1\): \(\sum_n \frac{1}{n}\) diverges.
Harmonic series.
4. Part (c) at \(z \in S^1 \setminus \theset{1}\): \(\sum_n \frac{z^n}{n}\) converges.
4.1. The partial sums \(A_N := \sum_{n=1}^{N} z^n\) are bounded.
Geometric series: \(A_N = \frac{z - z^{N+1}}{1 - z}\), so \(\abs{A_N} \leq \frac{2}{\abs{1 - z}}\), a constant independent of \(N\), since \(z \neq 1\).
4.2. The sequence \(\frac{1}{n}\) decreases to \(0\).
Monotone and bounded below by \(0\).
4.3. Dirichlet’s test applies: \(\sum \frac{z^n}{n}\) converges.
Dirichlet’s test: a series \(\sum a_n b_n\) converges if the partial sums of \(\sum a_n\) are bounded and \(b_n \downarrow 0\); take \(a_n = z^n\), \(b_n = 1/n\), using 1 and 2.
5. Q.E.D.
1 proves (a); 2 proves (b); 3 and 4 prove (c).