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Section 7.2 Series of Functions
Definition 7.16 .
Let \((f_n)_{n=1}^\infty\subseteq\mathcal{B}(X)\) for a set \(X\text{.}\) We'll say \(\sum_{n=1}^\infty f_n\) converges uniformly (pointwise) if \((\sum_{n=1}^N f_n)_{N=1}^\infty \subseteq\mathcal{B}(X)\) converges uniformly (pointwise).
Theorem 7.17 . Weierstrass M-test.
If \((f_n)_{n=1}^\infty\subseteq\mathcal{B}(X)\) for a set \(X\) and \(\sum_{n=1}^\infty \left|{\left|{{f_n}}\right|}\right|_\infty\) converges, then \(\sum_{n=1}^\infty f_n\) converges uniformly (or, equivalently, converges wrt \(d_\infty\) ).
Example 7.19 .
The series \(\sum_{n=1}^\infty \frac{1}{n^2+x^2}\) converges uniformly over \(x\in\mathbb{R}\text{.}\)
Solution . Let \(M_n=\frac{1}{n^2}\text{.}\) Then \(\left|{\frac{1}{n^2+x^2}}\right|\leq \frac{1}{n^2}\) for all \(x\in \mathbb{R}\) and \(\sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6}<\infty\text{,}\) so by the Weierstass \(M\) -test, \(\sum_{n=1}^\infty\frac{1}{n^2+x^2}\) converges uniformly over \(x\in \mathbb{R}\text{.}\)
Exploration 7.20 .
For any \(r>0\text{,}\) the series \(\sum_{n=0}^\infty \frac{z^n}{n!}\) converges uniformly on \(\{{z\in\mathbb{C}}\,:\,{\left|{z}\right|\leq r}\}\text{,}\) but \(\sum_{n=0}^\infty\frac{z^n}{n!}\) does not converge uniformly on \(\mathbb{C}\text{.}\)
