$\displaystyle \sum \lvert f_n \rvert$ converge uniformly $\implies$ $\displaystyle \sum f_n$ converge uniformly. If the series $\displaystyle \sum_{n=1}^\infty \lvert f_n \rvert$ converge uniformly then $\displaystyle \sum_{n=1}^\infty f_n$ converge uniformly?
is there some theorem proving that? I was thinking about some inequality or something like that, but I didn't see the solution. Can you help me?
 A: Define:
$$a_n(x)=\sum_{k=1}^n f_k(x)$$
Since $\sum_{k=1}^n |f_k(x)|$ is uniformly convergent,
$$\forall \epsilon, \exists N, \forall n,m>N, \forall x, |f_{n+1}(x)|+\cdots+|f_m(x)|\le\epsilon$$
We have:
$$|a_m(x)-a_n(x)|=|f_{n+1}(x)+\cdots+f_m(x)|\le |f_{n+1}(x)|+\cdots+|f_m(x)|\le\epsilon$$
So $a_n(x)$ is uniformly Cauchy, hence convergence uniformly.
Uniformly Cauchy iff Uniformly convergence.
A: Let $X$ denote the domain where the series $\sum_{j=1}^\infty \lvert f_j\rvert$ converges uniformly. Then
$$\sup_{x\in X}\left\lvert\sum_{j=1}^\infty f_j(x)-\sum_{j=1}^N f_j(x)\right\rvert=\sup_{x\in X}\left\lvert\sum_{j=N+1}^\infty f_j(x)\right\rvert\leq\sup_{x\in X}\sum_{j=N+1}^\infty \lvert f_j(x)\rvert\to 0$$
as $N\to\infty$, since $\sum_{j=1}^\infty \lvert f_j\rvert$ converges uniformly. It follows that $\sum_{j=1}^\infty  f_j$ converges uniformly.
A: First, I think the statement should be if $\sum_{n=1}^{k}|f_{n}|$ converges uniformly then $\sum_{n=1}^k\,f_{n}$ converges uniformly.
Otherwise there is no sequence to converge uniformly!
I don't think that you need Cauchy sequences! Define $g_{k}(x)=\sum_{n=1}^{k}|f_{n}(x)|$ and $g(x)=\sum_{n=1}^{+\infty}|f_{n}(x)|$. Therefore by uniform convergence $sup_{X}|g_{k}(x)-g(x)|\to  0$. Then
$sup_{X}|\sum_{n=k+1}^{+\infty}|f_{n}(x)||\to 0$ which implies:
$sup_{X}|\sum_{n=k+1}^{+\infty}f_{n}(x)|\to 0$ which gives :
$sup_{X}|\sum_{n=1}^{n=k}f_{n}(x)-\sum_{n=1}^{+\infty}f_{n}(x)|\to 0$
which is what we wanted to prove!
