Uniform convergence I need an example for sum $$\sum_{1}^{\infty } u_n(x)$$ which converge absolutely and uniformly in [a,b] while the sum $$\sum_{1}^{\infty } \left |u_n(x)  \right |$$ does not converge uniformly in [a,b].
 A: It would suffice to find a sequence of nonnegative functions $(u_n)$ and an interval $[a,b]$ such that 
$\ \ \ $1)  $\sum\limits_{n=0}^\infty u_n$ is convergent on $[a,b]$,
$\ \ \ $2) $\sum\limits_{n=0}^\infty u_n$ is not uniformly convergent on $[a,b]$, 
$\ \ \ $3) $(u_n)$ is a decreasing sequence of nonnegative functions.
and
$\ \ \ $4) $(u_n)$ converges uniformly to the zero function on $[a,b]$.  
For then the series $\sum\limits_{n=0}^\infty(-1)^n u_n(x)$ will converge uniformly  on $[a,b]$. 
Indeed, conditions 3) and 4)  guarantee that the series $\sum\limits_{n=0}^\infty(-1)^n u_n(x)$ is Uniformly Cauchy on $[a,b]$. To see this,
set $f_n(x)=(-1)^n u_n(x)$. Then for any $x\in[a,b]$ and for any two nonnegative integers $m$ and $n$ with $m\ge n$, we have by 3) that
$$
|f_n(x)+f_{n+1}(x)+\cdots+ f_m(x)|\le |f_n(x)|.
$$
Now from condition 4) we may make $|f_n(x)|$ uniformly small over $[a,b]$ by taking $n$ sufficiently large. 
   It follows that   $\sum\limits_{n=0}^\infty(-1)^n u_n(x)$ is Uniformly Cauchy, thus uniformly convergent, on $[a,b]$.
One example of   a sequence satisfying the conditions above is given in the comments. 
Another would be given by setting $u_n(x)={x^2\over (1+x^2)^n}$ on the interval $[-1,1]$. (Note, using standard facts of  Geometric series, that the pointwise limit of $(u_n)$ is $f(x)=\cases{1+x^2, & $x\ne0$\cr 0, &$x=0$}$. So $\sum\limits_{n=0}^\infty u_n$ converges on $[-1,1]$, but not uniformly there. Also note that by Bernoulli's inequality, for $x\ne0$ 
$${x^2\over(1+x^2)^n}\le {x^2\over 1+nx^2}={1\over n+(1/x^2)}\le {1\over n};$$ 
The above, together with the fact that  $u_n(0)=0$ for all $n$, shows that condition 4) is satisfied.)
