What is normal convergence?

Example 10.10 (One series, three verdicts)

Take $u_n(x) = \frac{x^n}{n}$ on $\intco{0}{1}$. Pointwise: converges for every $x \in \intco01$ (comparison with the geometric series). Normal on $\intcc{0}{a}$, $a < 1$: $\norm{u_n}_{\infty,\intcc0a} = \frac{a^n}{n}$, summable. Not normal on $\intco{0}{1}$: $\norm{u_n}_{\infty,\intco01} = \frac1n$, and $\sum\frac1n$ diverges. Not even uniform on $\intco{0}{1}$: the remainder resists near $1$,

so $\sup_{\intco01}\abs{R_N} \geq \frac12$ for every $N$. Closing insight: all four verdicts coexist peacefully — the sum $-\ln(1-x)$ is continuous on $\intco{0}{1}$ because continuity only needs uniformity near each point, i.e. on the segments $\intcc0a$; blowing up at the edge is the sum’s right.

Example 10.13 (The Riemann $\zeta$ function)

$\zeta(s) = \sum_{n\geq1} n^{-s}$ converges normally on every half-line $\intco{a}{+\infty}$, $a > 1$ ($\norm{n^{-s}}_\infty = n^{-a}$, summable): $\zeta$ is continuous on $\intoo{1}{+\infty}$; differentiating termwise (the derived series $\sum -\ln n\; n^{-s}$ also converges normally on $\intco{a}{\infty}$), $\zeta$ is $C^1$ — and, iterating, $C^\infty$ — with $\zeta'(s) = -\sum \frac{\ln n}{n^s}$. Note the discipline: normal convergence is checked on sub-half-lines, never on the open $\intoo{1}{\infty}$ itself, where it fails.

Example 10.14 (A logarithmic series, worked to the end)

Let $F(x) = \sum_{n\geq1} \frac{\eu^{-nx}}{n}$ on $\intoo{0}{\infty}$. Each term is bounded on $\intco{\delta} \infty$ by $\frac{\eu^{-n\delta}}{n} \leq \eu^{-n\delta}$, a convergent geometric series: normal convergence on every $\intco\delta\infty$, so $F$ is continuous on $\intoo{0}{\infty}$. The derived series $\sum -\eu^{-nx}$ is likewise normally convergent on $\intco\delta\infty$ ($\norm{\eu^{-nx}}_{\infty,\intco\delta\infty} = \eu^{-n\delta}$), so $F$ is $C^1$ with a geometric derivative:

Iterating, $F$ is $C^\infty$. Integrating $F'$ (both $F$ and $x \mapsto -\ln(1 - \eu^{-x})$ vanish at $+\infty$ and have the same derivative on $\intoo0\infty$):

the logarithmic series at $t = \eu^{-x}$. Closing insight: as $x \to 0^+$, $F(x) = -\ln(x + O(x^2)) = \ln\frac1x + O(x)$ — the series diverges logarithmically at the boundary, exactly like the harmonic series it becomes at $x = 0$; normal convergence on $\intco\delta\infty$ but not on $\intoo0\infty$ is the symptom.