# 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.

Title proof of Weierstrass’ criterion of uniform convergence ProofOfWeierstrassCriterionOfUniformConvergence 2013-03-22 16:26:28 2013-03-22 16:26:28 argerami (15454) argerami (15454) 4 argerami (15454) Proof msc 40A30 msc 26A15