absolute convergence in a Banach space কী?

Example 7.2

In $\mathcal{M}_n(K)$ (or $\mathcal{L}_c(E)$, $E$ Banach): for $\vertiii A < 1$, the Neumann series $\sum A^k$ converges absolutely to $(I - A)^{-1}$ (proved in Exercise 5.5); $\sum \frac{A^k}{k!}$ converges absolutely to $\eu^A$ for every $A$ (Example 5.22). Operator-valued geometric and exponential series behave like their scalar models — the whole point of the Banach framework.

Example 7.4

$\sum \frac{\sin n}{n}$ converges: $a_n = \frac1n \downarrow 0$ and $B_n = \sum_{k=1}^{n} \sin k$ is bounded — indeed $B_n = \Im\sum_{k \leq n} \eu^{\iu k} = \Im\,\frac{\eu^{\iu}(\eu^{\iu n} - 1)}{\eu^{\iu} - 1}$, of modulus $\leq \frac{2}{\abs{\eu^{\iu} - 1}}$. It does not converge absolutely ($\abs{\sin n} \geq \sin^2 n = \frac{1 - \cos 2n}{2}$, and $\sum \frac{1 - \cos 2n}{2n}$ diverges since $\sum \frac{\cos 2n}{n}$ converges by the same Abel test while $\sum \frac{1}{2n}$ diverges). The alternating series test is the special case $b_n = (-1)^n$.

Example 7.5 (Abel on the circle of convergence)

For which complex $z$ with $\abs z = 1$ does $\sum_{n \geq 1} \frac{z^n}{n}$ converge? At $z = 1$ it is the harmonic series: divergent. For $z \neq 1$ on the circle, Abel’s test applies with $a_n = \frac1n \downarrow 0$ and $b_n = z^n$, whose partial sums are bounded independently of $N$:

Convergent — though never absolutely ($\sum\frac1n$). One series, a circle of behaviours: divergence at a single point, semi-convergence everywhere else. This is the standard boundary behaviour of power series (Chapter 11), met here with bare hands; at $z = -1$ it recovers the alternating harmonic series, and at $z = \eu^{\iu\theta}$ its real and imaginary parts are the series $\sum\frac{\cos n\theta}{n}$ and $\sum\frac{\sin n\theta}{n}$ of Exercise 7.4.