Let $R$ be an open subset of $\mathbb{R}$ and $f_n:R\to\mathbb{R}$, and assume that $F(x):=\sum_{n} f_n(x)$ converges uniformly on $R$ because of Weierstrass' test, i.e. $|f_n(x)|\le\alpha_n$ for every $x\in R$ and $\sum_n \alpha_n<\infty$.
Let's introduce some weights $\lambda_n\in(0,1]$ and consider $\Lambda(x):=\sum_{n}\lambda_n f_n(x)$. Obviously $|\lambda_n f_n(x)|\le\alpha_n$, so $\Lambda(x)$ is uniformly convergent over $R$ by Weierstrass' test again.
Can we say that $|\Lambda(x)|\le|F(x)|$ for every $x\in R$?
Surely we have $|\Lambda(x)|\le\sum_n |\lambda_n f_n(x)|\le \sum_n |f_n(x)|$, but then I don't see how to relate the last sum to $F(x)$ other than the opposite inequality ($\ge$)!
I suspect that is probably not the case, but I really cannot see any argument to prove or disprove it. Any idea/hint/result I might not be aware of?
Take $f_1=4$, $f_2=-4$ and $f_n=0$ for $n \geq 3$. And choose $\lambda_1=1/2,\lambda=\lambda_2=\frac{1}{4}$.
Then $F(x)= 4-4=0$ everywhere, and $\Lambda(x)=\frac{1}{2}\cdot 4-\frac{1}{4}\cdot 4=1$.
Therefore, $|F(x)|<|\Lambda (x)|$ everywhere.
You need $f_n$ to be positive for the result to hold for any choise of $\lambda_i$, also notice that the reverse holds, as well.