Uniform convergence on the compact metric space Let $X$ be a compact metric space and let $f_n$ and $g_n$ be sequences of continuous functions from the metric space into the complex plane. If $\Sigma g_n$ converges uniformly and absolutely on $X$ and the absolute value of $f_n$ is equal or smaller than the absolute value of $g_n$ for each $n$, then, it is clear that $\Sigma f_n$ converges absolutely on $X$. However how can I show that $\Sigma f_n$ converges uniformly on $X$? I tried the Weierstrass M-test but failed... Could anyone please help me?
 A: This is wrong. It assumes that $\sum |g_n|$ converges uniformly, which does not follow from the assumptions.
Let $\varepsilon > 0$.
Since $\sum_{n=1}^\infty |g_n|$ converges uniformly, there exists $n_0 \in \mathbb{N}$ such that
$$n \ge n_0 \implies \forall x\in X \text{ holds }\sum_{k=n+1}^\infty |g_k(x)| = \left|\sum_{k=1}^\infty |g_k(x)|- \sum_{k=1}^n|g_k(x)|\right| < \varepsilon$$
Now for $M, N \ge n_0$ and $x \in X$ we have:
$$\left|\sum_{k=1}^M f_k(x)- \sum_{k=1}^N f_k(x)\right|= \left|\sum_{k=N+1}^M f_k(x)\right| \le \sum_{k=N+1}^M |f_k(x)| \le \sum_{k=N+1}^M |g_k(x)| \le \sum_{k=N+1}^\infty |g_k(x)| < \varepsilon$$
Therefore, $\sum_{k=1}^n f_k$ is uniformly Cauchy. This means that $\sum_{k=1}^n f_k$ is Cauchy in $C(X)$, the space of all continuous functions from $X$ to $\mathbb{C}$, equipped with the uniform metric: $$d_\infty(f_1, f_2) = \sup_{x \in X} |f_1(x) - f_2(x)|$$
Since $(C(X), d_\infty)$ is a complete metric space, the series $\sum_{k=1}^n f_k$ converges with respect to $d_\infty$, which in turn implies that $\sum_{k=1}^n f_k$ converges absolutely.
