proof of Weierstrass’ criterion of uniform convergence

The assumption that $|f_{n}(x)|\\leq M_{n}$ for every $x$ guarantees that each numerical series $\\sum_{n}f_{n}(x)$ converges absolutely. We call the limit $f(x)$ .

To see that the convergence is uniform: let $\\epsilon>0$ . Then there exists $K$ such that $n>K$ implies $\\sum_{n>K}M_{n}<\\epsilon$ . Now, if $k>K$ ,

|f(x)-\\sum_{n=1}^{k}f_{n}(x)|=|\\sum_{n>k}f_{n}(x)|\\leq\\sum_{n>k}|f_{n}(x)|\\leq%\\sum_{n>k}M_{n}<\\epsilon

The $\\epsilon$ does not depend on $x$ , so the convergence is uniform.