[proofplan]
We establish the logical independence by constructing two counterexamples. For part (1), the geometric series $\sum x^n$ on $(-1,1)$ converges absolutely at every point but fails to converge uniformly because its tails blow up near $x = 1$. For part (2), the alternating harmonic series $\sum (-1)^{n-1}/n$, viewed as constant functions, converges uniformly (the supremum over $E$ equals the pointwise value) but diverges absolutely since $\sum 1/n = \infty$.
[/proofplan]
[step:Exhibit a series that converges absolutely but not uniformly]
Define $g_n: (-1,1) \to \mathbb{R}$ by $g_n(x) = x^n$. For each fixed $x \in (-1,1)$, the series of absolute values satisfies $\sum_{n=0}^\infty |x|^n = 1/(1-|x|) < \infty$, so $\sum g_n$ converges absolutely at every point of $(-1,1)$.
We show the convergence is not uniform. For each $N \in \mathbb{N}$, the partial-sum remainder on $(0,1)$ is
\begin{align*}
R_N(x) = \sum_{n=N+1}^\infty x^n = \frac{x^{N+1}}{1 - x}, \quad x \in (0,1).
\end{align*}
As $x \to 1^-$, the denominator $1 - x \to 0$ while $x^{N+1} \to 1$, so $R_N(x) \to +\infty$. Therefore
\begin{align*}
\sup_{x \in (-1,1)} |R_N(x)| \geq \sup_{x \in (0,1)} \frac{x^{N+1}}{1 - x} = +\infty
\end{align*}
for every $N$, and the convergence is not uniform on $(-1,1)$.
[guided]
We need a series that converges absolutely at every point but fails to converge uniformly. The geometric series is the simplest candidate: its sum is known in closed form, making the tail easy to analyse.
Define $g_n: (-1,1) \to \mathbb{R}$ by $g_n(x) = x^n$. For each fixed $x \in (-1,1)$, we have $|x| < 1$, so the geometric series of absolute values converges:
\begin{align*}
\sum_{n=0}^\infty |g_n(x)| = \sum_{n=0}^\infty |x|^n = \frac{1}{1 - |x|} < \infty.
\end{align*}
This establishes absolute convergence at every point.
Why does uniform convergence fail? The issue is that the rate of convergence deteriorates as $x$ approaches the boundary of the interval. For $x \in (0,1)$, the closed-form expression for the partial-sum remainder is
\begin{align*}
R_N(x) = \sum_{n=N+1}^\infty x^n = \frac{x^{N+1}}{1 - x}.
\end{align*}
The factor $1/(1-x)$ blows up as $x \to 1^-$, while $x^{N+1} \to 1$. Therefore
\begin{align*}
\sup_{x \in (-1,1)} |R_N(x)| \geq \sup_{x \in (0,1)} \frac{x^{N+1}}{1 - x} = +\infty
\end{align*}
for every fixed $N$. Since the supremum of the remainder never tends to zero, the convergence is not uniform. The root cause is the lack of compactness: on any closed sub-interval $[-r, r]$ with $r < 1$, the convergence is uniform, but on the full open interval it is not.
[/guided]
[/step]
[step:Exhibit a series that converges uniformly but not absolutely]
Define $g_n: E \to \mathbb{R}$ by $g_n(x) = (-1)^{n-1}/n$ for all $x \in E$, where $E$ is any non-empty set. Each $g_n$ is a constant function, so the partial sums $S_N(x) = \sum_{n=1}^N (-1)^{n-1}/n$ are independent of $x$.
By the [alternating series test](/theorems/177), the numerical series $\sum_{n=1}^\infty (-1)^{n-1}/n$ converges to $\ln 2$. Since the partial sums do not depend on $x$, the supremum of the remainder satisfies
\begin{align*}
\sup_{x \in E} |S_N(x) - \ln 2| = |S_N - \ln 2| \to 0 \quad \text{as } N \to \infty,
\end{align*}
so the convergence is uniform on $E$.
However, the series of absolute values is the harmonic series $\sum_{n=1}^\infty 1/n = +\infty$, so the convergence is not absolute at any point of $E$.
[guided]
We need a series that converges uniformly but fails to converge absolutely. The simplest construction uses constant functions: if each $g_n$ takes the same value at every point, then uniform and pointwise convergence coincide, and we only need to find a conditionally convergent numerical series.
Define $g_n: E \to \mathbb{R}$ by $g_n(x) = (-1)^{n-1}/n$ for all $x \in E$, where $E$ is any non-empty set. Each $g_n$ is constant, so the partial sums $S_N(x) = \sum_{n=1}^N (-1)^{n-1}/n$ do not depend on $x$ at all.
The [alternating series test](/theorems/177) applies: the sequence $1/n$ is monotonically decreasing to zero, and the signs alternate. Therefore the numerical series converges, with limit $\ln 2$. Since the convergence is purely numerical (no $x$-dependence), the supremum of the error is
\begin{align*}
\sup_{x \in E} |S_N(x) - \ln 2| = |S_N - \ln 2| \to 0,
\end{align*}
which gives uniform convergence on $E$.
For absolute convergence, we examine the series of absolute values:
\begin{align*}
\sum_{n=1}^\infty |g_n(x)| = \sum_{n=1}^\infty \frac{1}{n} = +\infty.
\end{align*}
This is the harmonic series, which diverges. So $\sum g_n$ does not converge absolutely at any point, completing the proof that uniform convergence does not imply absolute convergence.
[/guided]
[/step]
