Prove: If $\sum \limits_{n=1}^{\infty}|f_n|$ converges uniformly, so does $\sum \limits_{n=1}^{\infty}f_n$. I'd like your help proving that 

If $\sum\limits_{n=1}^{\infty}|f_n|$ converges uniformly, so does $ \sum\limits_{n=1}^{\infty}f_n$.

There a Weierstrass theorem saying that if there's a positive sum $ \sum\limits_{k=1}^{\infty}b_k$ which converges and $|u_k(x)|\leq b_k$ for every $x$ so $ \sum\limits_{k=1}^{\infty}u_k$ converges uniformly. Can it goes the opposite way too? And then we can claim that there $ \sum\limits_{k=1}^{\infty}b_k$ such that $\|f_n\|\leq b_k$ and so does $|f_n|\leq b_k$ and we are done? 
If not, we know that in general if $ \sum\limits_{k=1}^{\infty}|b_k|$ converges, so does $ \sum\limits_{k=1}^{\infty}b_k$, how can apply the uniform convergence here?
Thanks a lot! 
 A: I don't think the converse of the Weierstrass test holds. However, you can use the following 
Fact: A series $\sum\limits_{i=1}^\infty f_i(x)$ of real-valued functions converges uniformly on a set $E$ if and only if it is uniformly Cauchy.  That is, if and only if, given $\epsilon>0$, there is an $N$ such that $$\Bigl|\sum_{i=n}^m f_i(x)\,\Bigr |<\epsilon$$ for all $m\ge n\ge N$ and for all   $x\in E$.
Now think about this and the triangle inequality.

$\color{maroon}{\text {Warning! Solution follows:}}$

We now show that your series $\sum\limits_{i=n}^m  f_i(x)$ converges uniformly:
Let $\epsilon>0$.
Since,  $\sum\limits_{i=1}^\infty |f_i(x)|$ converges uniformly, there is an $N$ such that for all $m\ge n\ge N$ and for all $x$:
$$ \sum_{i=n}^m |f_i(x)|  <\epsilon.$$
Using this and the triangle inequality, it follows that  for all $m\ge n\ge N$ and for all $x$:
$$
 \Bigl|\sum_{i=n}^m  f_i(x)\,\Bigr| \le \sum_{i=n}^m |f_i(x)|  <\epsilon.
$$
So $\sum\limits_{i=1}^\infty  f_i(x)$ is uniformly Cauchy, and thus uniformly convergent.

For completeness:

The Fact above follows (by looking at the sequence of partial sums of the series) from the following standard theorem:
Theorem:
A sequence of real-valued functions $\{f_n\}$ is uniformly convergent on a set $E$ if and only if it is uniformly Cauchy on $E$; that is, given $\epsilon>0$, there is an $N$ so that
$$
\tag{1}|f_n(x)-f_m(x)|<\epsilon,\quad  \text{ for all }n,m\ge N\text{  and for all }x\in E
$$
Proof: 
To prove the forward implication, suppose $\{f_n\}$ converges uniformly to $f$ on the set $E$. Then, given $\epsilon>0$,
there is an $N$ so that 
$$
|f_n(x)-f(x)|<\epsilon/2
$$
for all $n\ge N$ and for all $x\in   E$.
Thus, if $m,n\ge N$ and $x\in E$
 $$
|f_n(x)-f_m(x)|<|f_n(x)-f(x)|+|f_m(x)-f(x)|<{\epsilon\over2}+{\epsilon\over2}=\epsilon.
$$
From this, it follows that $\{f_n\} $ is uniformly Cauchy on $E$.
To prove the reverse implication, suppose $\{f_n\}$ is uniformly Cauchy on $E$. Then for each $x\in E$, the sequence $\{f_n(x)\}$ is Cauchy and thus converges to some number $f(x)$.
We claim that $\{f_n\}$ converges uniformly to $f$, as defined above, on $E$.
Towards proving the claim,
let $\epsilon>0$ and choose $N$ a positive integer that verifies equation (1).
Then if $m\ge N$ is fixed,  $n\ge N$, and $x\in E$:
$$
\tag{2}|f_m(x)-f_n(x)|<\epsilon  
$$
Taking the limit as $n\rightarrow\infty$ in (2) gives
$$
\tag{3}|f_m(x)-f (x)|\le\epsilon . 
$$
Since $\epsilon$ was an arbitrary positive number, and since (3) holds for all $m\ge N$ and all $x\in E$, it follows that $\{f_n\}$ converges uniformly to $f$ on $E$.
