Why normal convergence $\implies $ uniform convergence.

Question

We say that $\sum_{k=0}^\infty f_k(x)$ converge normally in $I$ if there is $M_n$ s.t. $|f_n(x)|\leq M_n$ for all $n$ and all $x\in I$ with $\sum_{k=0}^\infty M_n$ converge. Why does it implies that the serie converge uniformly ? I set $f(x)=\sum_{k=0}^\infty f_k(x)$. Why $$\sup_{I}|f(x)-\sum_{k=0}^n f_k(x)|\to 0$$ when $n\to \infty $ ?

Answer 1

$$\left|f(x)-\sum_{k=0}^n f_k(x)\right|=\left|\sum_{k=0}^\infty f_k(x)-\sum_{k=0}^n f_k(x)\right|=\left|\sum_{k={n+1}}^\infty f_k(x) \right|\leq \sum_{k=n+1}^\infty M_n,$$ and thus $$\sup_{x\in I}\left|f(x)-\sum_{k=0}^n f_k(x)\right|\leq \sum_{k=n+1}^\infty M_n\underset{n\to \infty }{\longrightarrow }0.$$

Answer 2

It converges uniformly (and absolutely) because it satisfies the uniform Cauchy criterion:

if $N$ is such that $$|M_n+M_{n+1}+\dots +M_p|<ε \qquad\text{for all }\: p>n\ge N,$$ we also have for all $x\in I$ \begin{align} |f_n(x)+f_{n+1}(x)+\dots +f_p(x)|&\le |f_n(x)|+|f_{n+1}(x)|+\dots +|f_p(x)|\\ &\le|M_n+M_{n+1}+\dots +M_p|< ε. \end{align}