[proofplan]
We exhibit the series $\sum_{n=1}^\infty (-1)^n x^n/n$ on $E = [0,1)$ as a counterexample. Pointwise absolute convergence follows from comparison with the geometric series. Uniform convergence of the original series follows from the alternating series estimation, which gives a remainder bound of $1/(N+1)$ independent of $x$. Absolute uniform convergence fails because the tails of $\sum x^n/n$ approach partial sums of the harmonic series as $x \to 1^-$.
[/proofplan]
[step:Verify pointwise absolute convergence of $\sum (-1)^n x^n/n$ on $[0,1)$]
Fix $x \in [0,1)$. The series of absolute values satisfies
\begin{align*}
\sum_{n=1}^\infty \left|\frac{(-1)^n x^n}{n}\right| = \sum_{n=1}^\infty \frac{x^n}{n}.
\end{align*}
Since $0 \leq x < 1$, we have $x^n/n \leq x^n$ for all $n \geq 1$, and $\sum_{n=1}^\infty x^n = x/(1-x) < \infty$. By the comparison test, $\sum x^n/n$ converges. Since $x \in [0,1)$ was arbitrary, the series converges absolutely at every point of $[0,1)$.
[/step]
[step:Establish uniform convergence of $\sum (-1)^n x^n/n$ on $[0,1)$ via alternating series estimation]
For each fixed $x \in [0,1)$, the sequence $a_n(x) = x^n/n$ is decreasing (since $a_{n+1}(x)/a_n(x) = xn/(n+1) \leq x < 1$ for $n \geq 1$) and converges to zero. By the alternating series estimation, the remainder after $N$ terms satisfies
\begin{align*}
|R_N(x)| = \left|\sum_{n=N+1}^\infty \frac{(-1)^n x^n}{n}\right| \leq \frac{x^{N+1}}{N+1} \leq \frac{1}{N+1}
\end{align*}
for all $x \in [0,1)$. Taking the supremum over $x$:
\begin{align*}
\sup_{x \in [0,1)} |R_N(x)| \leq \frac{1}{N+1} \to 0 \quad \text{as } N \to \infty.
\end{align*}
Therefore $\sum (-1)^n x^n/n$ converges uniformly on $[0,1)$.
[guided]
We need to show that the alternating series converges uniformly, even though the interval $[0,1)$ is not compact. The key is that the alternating series estimation provides a bound on the remainder that does not depend on $x$.
For each fixed $x \in [0,1)$, consider the sequence $a_n(x) = x^n/n$. We verify the hypotheses of the alternating series estimation: the ratio $a_{n+1}(x)/a_n(x) = xn/(n+1)$ satisfies $xn/(n+1) \leq x < 1$ for all $n \geq 1$, so the sequence is decreasing. Moreover $a_n(x) = x^n/n \to 0$ as $n \to \infty$. The alternating series estimation then gives
\begin{align*}
|R_N(x)| \leq a_{N+1}(x) = \frac{x^{N+1}}{N+1}.
\end{align*}
Since $x \in [0,1)$, we have $x^{N+1} < 1$, so $|R_N(x)| \leq 1/(N+1)$ for all $x \in [0,1)$. This bound is independent of $x$, which is the essential point:
\begin{align*}
\sup_{x \in [0,1)} |R_N(x)| \leq \frac{1}{N+1} \to 0.
\end{align*}
The uniform convergence follows despite the non-compactness of $[0,1)$, because the alternating structure provides cancellation that tames the tails uniformly.
[/guided]
[/step]
[step:Show that $\sum x^n/n$ does not converge uniformly on $[0,1)$]
For any $M < N$, the partial tail of the series of absolute values satisfies
\begin{align*}
\sup_{x \in [0,1)} \sum_{j=M+1}^{N} \frac{x^j}{j} \geq \lim_{x \to 1^-} \sum_{j=M+1}^{N} \frac{x^j}{j} = \sum_{j=M+1}^{N} \frac{1}{j}.
\end{align*}
The right-hand side is a partial sum of the harmonic series. For any fixed $M$ and any $\varepsilon > 0$, choosing $N$ large enough makes $\sum_{j=M+1}^{N} 1/j > \varepsilon$. Therefore the partial sums of $\sum x^n/n$ are not uniformly Cauchy on $[0,1)$, and the series $\sum |g_n|$ does not converge uniformly. This completes the proof that uniform convergence plus pointwise absolute convergence does not imply absolute uniform convergence.
[guided]
We have shown that $\sum (-1)^n x^n/n$ converges both uniformly and absolutely (pointwise) on $[0,1)$. It remains to show that the series of absolute values $\sum x^n/n$ fails to converge uniformly.
The idea is that as $x \to 1^-$, the tails of $\sum x^n/n$ approach the tails of the harmonic series $\sum 1/n$, which diverge. Concretely, for any integers $M < N$:
\begin{align*}
\sup_{x \in [0,1)} \sum_{j=M+1}^{N} \frac{x^j}{j} \geq \lim_{x \to 1^-} \sum_{j=M+1}^{N} \frac{x^j}{j} = \sum_{j=M+1}^{N} \frac{1}{j}.
\end{align*}
The passage to the limit is valid because each $x^j/j$ is continuous on $[0,1]$ and the sum is finite. Now $\sum_{j=M+1}^{N} 1/j$ is a partial sum of the harmonic series, which can be made arbitrarily large by choosing $N$ sufficiently large relative to $M$. Therefore no $K$ exists such that $\sup_{x \in [0,1)} \sum_{j=K+1}^{N} x^j/j < \varepsilon$ for all $N > K$, and the series of absolute values does not converge uniformly.
The underlying reason is the same as in the geometric series example: the lack of compactness of $[0,1)$ allows the tails to blow up near the missing endpoint. The alternating signs in $\sum (-1)^n x^n/n$ provide cancellation that keeps the original series uniformly convergent, but no such cancellation is available in $\sum x^n/n$.
[/guided]
[/step]
