I was reading about the Weierstrass $M$-test, and I was wondering about the following:
Suppose that $(f_{n})$ is a sequence of real-valued functions defined on a set $A$, and there exists $(g_{n})_{n \in \mathbb{N}} \geq 0$ defined on $A$ such that $\sum_{i=1}^{\infty}g_{i}$ is uniformly convergent on $A$, and for all $n\in \mathbb{N}, x \in A$, $|f_{n}(x)|\leq g_{n}(x)$. Then does $\sum_{i=1}^{\infty}f_i$ converge absolutely and uniformly on $A$?
I think this is true by a slight modification of the original proof. However, I cannot find anything online relating to this. So is true?
It's true by checking the Cauchy criterion for the partial sums of the series $\sum f_n(x)$, the same way as is done for the $M$-test. Since $\sum g_n(x)$ converges uniformly for $x\in A$, for each $\epsilon > 0$, there exists $N(\epsilon)$ such that if $p,q\ge N$, for all $x\in A$, $$ |\sum_{n=p}^qf_n(x)| \le\sum_{n=p}^q|f_n(x)|\le \sum_{n=p}^q g_n(x) < \epsilon. $$ This implies that the partial sums of the series $\sum |f_n(x)|$ are uniformly Cauchy, so $\sum f_n(x)$ is uniformly and absolutely convergent in $x\in A$.