Weierstrass M-test with functions instead of numbers as bounds Consider a set $A \subseteq \mathbb R$ and a sequence of functions $f_n: A \to \mathbb R$. The Weierstrass M-test for uniform convergence of $\sum_{n=1}^\infty f_n(x)$ states:
Theorem Assume that there is a sequence of numbers $M_n$ such that $|f_n(x)|\leq M_n$ for all $n$ and for all $x \in A$, and that $\sum_{n=1}^\infty M_n < \infty$. Then the series $\sum_{n=1}^\infty f_n(x)$ converges absolutely and uniformly on $A$.
Does the uniform convergence result still hold if the series of numbers $M_n$ is replaced by a series of functions $g_n$ that converges uniformly? That is, is the following proposition true? In my application the functions $f_n$ are positive, in case it helps.
Proposition Assume that there is a sequence of functions $g_n: A \to \mathbb R$ such that $|f_n(x)|\leq g_n(x)$ for all $n$ and for all $x \in A$, and that the series $\sum_{n=1}^\infty g_n(x)$ converges uniformly on $A$. Then $\sum_{n=1}^\infty f_n(x)$ converges uniformly on $A$.
 A: Yes, it is true.
Take $\varepsilon>0$. Since the series $\sum_{n=1}^\infty g_n(x)$ converges, there is a $p\in\mathbb N$ such that$$(\forall x\in A)(\forall m,n\in\mathbb{N}):m\geqslant n\geqslant p\implies\sum_{k=n}^mg_n(x)<\varepsilon.$$But then$$(\forall x\in A)(\forall m,n\in\mathbb{N}):m\geqslant n\geqslant p\implies\sum_{k=n}^m\bigl|f_n(x)\bigr|<\varepsilon,$$and therefore$$(\forall x\in A)(\forall m,n\in\mathbb{N}):m\geqslant n\geqslant p\implies\left|\sum_{k=n}^mf_n(x)\right|<\varepsilon,$$which means that the series $\sum_{n=1}^\infty f_n(x)$ converges uniformly.
A: Let $x \in A$. Now, since 
$$
\sum_{i \geq 1}f_i(x) \leq \sum_{i\geq 1}g_i(x) < \infty
$$
the function $f := \sum_{i \geq 1}f_i$ exists and $\sum\limits_{1 \leq i \leq n}f_i$ converges pointwise to $f$. Let $g = \sum_{i\geq 1}g_i$.To see uniform convergence, note that
$$
\|f-\sum\limits_{1 \leq i \leq n}f_i\|_\infty = \sup_{x\in A}\left|\sum_{i \geq n+1} f_i(x)\right| \leq \sup_{x \in A} \sum_{i \geq n+1}|f_i(x)|
\leq \sup_{x \in A}\sum_{i \geq n+1}g_i(x) \leq  \\ \leq \sup_{x\in A}\left|\sum_{i \geq n+1}g_i(x)\right| = \sup_{x\in A}\left|g-\sum_{1 \leq i \leq n}g_i(x)\right| = \|g - \sum_{1 \leq i \leq n}g_i(x)\|_\infty \xrightarrow{n \to \infty} 0
$$
because $\sum_{i \geq 1}g_i \to g$ uniformly.
A: Hint: For any $m <n,$
$$|\sum_{k=m}^{n}f_k(x)| \le \sum_{k=m}^{n}|f_k(x)| \le \sum_{k=m}^{n}g_k(x).$$
