Uniform convergence, but no absolute uniform convergence

Solution 1:

Take $f_n(x)=(-1)^n {x^n\over n}$, for $n\ge 1$ on $[0,1)$.

Then if $n$ and $m$ are positive integers with $m\ge n$ and $x\in[0,1)$: $$ |f_n(x) +f_{n+1}(x)+\cdots +f_m(x)|\le| f_n(x)| ={x^n\over n}\le {1\over n}. $$ This implies that $\sum\limits_{n=1}^\infty f_n(x)$ converges uniformly on $[0,1)$.

The series $\sum\limits_{n=1}^\infty |f_n(x)|$ converges on $[0,1)$ as comparison with the series $\sum\limits_{n=1}^\infty x^n$ will show.

But $\sum\limits_{n=1}^\infty |f_n(x)|$ does not converge uniformly on $[0,1)$; since, for any $n$, $$ |f_n(x)| +|f_{n+1}(x)|+\cdots +|f_{2n}(x)| \ge {1\over 2n} \cdot nx^{2n}, $$ and $\lim\limits_{x\rightarrow1^-} x^{2n} =1$.

I think that any absolutely convergent but non-uniformly convergent series of positive, decreasing terms $f_n$ with $(f_n)$ converging uniformly to 0 would provide a counterexample by making the series "alternating".