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Uniform convergence
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===To series=== We say that <math display="inline">\sum_{n=1}^\infty f_n</math> converges: {{ordered list | list-style-type=lower-roman | pointwise on ''E'' if and only if the sequence of partial sums <math>s_n(x)=\sum_{j=1}^{n} f_j(x)</math> converges for every <math>x\in E</math>. | uniformly on ''E'' if and only if ''s''<sub>''n''</sub> converges uniformly as <math>n\to\infty</math>. | absolutely on ''E'' if and only if <math display="inline">\sum_{n=1}^\infty |f_n|</math> converges for every <math>x \in E</math>. }} With this definition comes the following result: <blockquote>Let ''x''<sub>0</sub> be contained in the set ''E'' and each ''f''<sub>''n''</sub> be continuous at ''x''<sub>0</sub>. If <math display="inline"> f = \sum_{n=1}^\infty f_n</math> converges uniformly on ''E'' then ''f'' is continuous at ''x''<sub>0</sub> in ''E''. Suppose that <math>E = [a, b]</math> and each ''f''<sub>''n''</sub> is integrable on ''E''. If <math display="inline">\sum_{n=1}^\infty f_n</math> converges uniformly on ''E'' then ''f'' is integrable on ''E'' and the series of integrals of ''f''<sub>''n''</sub> is equal to integral of the series of f<sub>n</sub>.</blockquote>
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