TITLE: Uniform convergence, but no absolute uniform convergence
QUESTION [8 upvotes]: Can someone give an example of a series of functions $f_k(x)$ for which
$\sum_{k=0}^{\infty} f_k(x)$ converges uniformly, and $\sum_{k=0}^{\infty} |f_k(x)|$ converges pointwise, but $\sum_{k=0}^{\infty} |f_k(x)|$ does not converge uniformly.
I'm having trouble finding an example (or proving this is impossible), as there is no certainty the limit of the absolute and relative series are equal.

REPLY [12 votes]: 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".