The power series are \[\sum_{n=0}^{\infty}c_nx^n\] The numbers \(c_n\) are called coefficients. \[R=\frac{1}{\displaystyle\lim_{n\to \infty}\text{sup}\sqrt[n]{|c_n|}}\] is called the radius of convergence of power series \[\sum_{n=0}^{\infty}c_nx^n\] \[\displaystyle\lim_{n\to \infty}\text{sup}\sqrt[n]{|c_nx^n|}=\frac{|x|}{R}\] Then \[\sum_{n=0}^{\infty}c_nx^n\] converges if \(|x|<R\) and diverges if \(|x|>R\).
Suppose the series \(\sum_{n=0}^{\infty}c_nx^n\) converges for \(|x|<R\), then it converges uniformly on \([-R+\varepsilon,R-\varepsilon], \varepsilon>0\). For \(|x|\leq R-\varepsilon\) we have \[|c_nx^n|\leq|c_n(R-\varepsilon)^n|\] and since \[\sum c_n(R-\varepsilon)^n\] converges absolutely (every power series converges absolutely in the interior of its radius), then there is an integer \(N\) that \[|\sum_{i=0}^{n}c_ix^i-\sum_{i=0}^{m}c_ix^i|=|\sum_{m+1}^{n}c_ix^i|<\varepsilon,\quad n\ge m\ge N\] then \(\sum_{n=0}^{\infty}c_nx^n\) converges uniformly.
Functions which are presented by power series \[f(x)=\sum_{n=0}^{\infty}c_nx^n\] or \[f(x)=\sum_{n=0}^{\infty}c_n(x-a)^n\] are called analytic functions. The function \[f(x)=\sum_{n=0}^{\infty}c_nx^n\] is continuous and differentiable in \((-R,R)\).
If \(\sum c_n\) converges, the limit of analytic function \[\lim_{x\to 1}f(x)=\lim_{x\to 1}\sum_{n=0}^{\infty}c_nx^n=\sum_{n=0}^{\infty}c_n\quad (-1<x<1)\] Let \[s_n=\sum_{i=0}^{n}c_i\] Then \[\sum_{n=0}^{m}c_nx^n=\sum_{n=0}^{m}(s_n-s_{n-1})x^n=\sum_{n=0}^{m}s_nx^n-\sum_{n=0}^{m}s_{n-1}x^n=\sum_{n=0}^{m-1}s_nx^n+s_mx^m-x\sum_{n=0}^{m-1}s_nx^n\\ =(1-x)\sum_{n=0}^{m-1}s_nx^n+s_mx^m\] For \(|x|<1\), let \(m\to \infty\) and obtain \[f(x)=(1-x)\sum_{n=0}^{\infty}s_nx^n\] Suppose \[s=\lim_{n\to\infty}s_n\] let \(\varepsilon>0\), because \(s_n=\sum_{i=0}^{n}c_i\) converges, choose \(N\) so that \(n>N\) implies \[|s-s_n|<\frac{\varepsilon}{2}\] then since \[(1-x)\sum_{n=0}^{\infty}x^n=1\quad(|x|<1)\] we obtain \[|f(x)-s|=|(1-x)\sum_{n=0}^{\infty}(s_n-s)x^n|\leq (1-x)\sum_{n=0}^{N}|s_n-s||x|^n+\frac{\varepsilon}{2}\leq \varepsilon\] Because \(x\to1\) we can choose suitable \(x\) to make \[(1-x)\sum_{n=0}^{N}|s_n-s||x|^n\leq \frac{\varepsilon}{2}\]
If \(\sum a_n\) is a series of complex numbers which converges absolutely, then every rearrangement of \(\sum a_n\) converges and they all converge to the same sum. Let \(\sum a'_n\) be a rearrangement, with partial sums \(s'_n\). Given \(\varepsilon>0\), because \(\sum a_n\) converges absolutely, there exists an integer \(N\) such that \(m\ge n\ge N\) implies \[\sum_{i=n}^{m}|a_i|\le\varepsilon\] Now choose \(p\) such that the integers \(1,2,\cdots,N\) are all contained in the set \(k_1,k_2,\cdots,k_p\), then if \(n>p\), the numbers \(a_1,a_2,\cdots,a_N\) will cancel in the difference \(s_n-s'_n\), so that \[|s_n-s'_n|\leq\varepsilon\] Hence \(\{s'_n\}\) converges to the same sum as \(\{s_n\}\).
The double sequence \(\{a_{ij}\},i=1,2,3\cdots,j=1,2,3,\cdots\) suppose that \[\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}|a_{ij}|\] converges, Then the different rearrangements of \(\{a_{ij}\}\) converge to the same sum \[\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}a_{ij}=\sum_{j=1}^{\infty}\sum_{i=1}^{\infty}a_{ij}\]
Suppose \[f(x)=\sum_{n=0}^{\infty}c_nx^n\] the series converging in \(|x|<R\) or \[\lim_{n\to \infty}\text{sup}\sqrt[n]{|c_nx^n|}<1\] We have \[\begin{align} f(x)&=\sum_{n=0}^{\infty}c_nx^n\\ &=\sum_{n=0}^{\infty}c_n(x-a+a)^n\\ &=\sum_{n=0}^{\infty}c_n\sum_{m=0}^{n}{n \choose m}(x-a)^ma^{n-m}\\ \end{align}\] Because if \(|x-a|+|a|<R\) \[\sum_{n=0}^{\infty}\sum_{m=0}^{n}\Biggl|c_n{n \choose m}(x-a)^ma^{n-m}\Biggr|=\sum_{n=0}^{\infty}|c_n|\cdot(|x-a|+|a|)^n\] converges and when \(n\to \infty, n\ge m\) \[\begin{align} f(x)&=\sum_{n=0}^{\infty}c_n\sum_{m=0}^{n}{n \choose m}(x-a)^ma^{n-m}\\ &=\sum_{m=0}^{\infty}\Biggl[\sum_{n=m}^{\infty}{n \choose m}c_na^{n-m}\Biggr](x-a)^m\\ &=\sum_{m=0}^{\infty}\Biggl[\frac{f^{(m)}(a)}{m!}\Biggr](x-a)^m\\ &=\sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n \quad(|x-a|+|a|<R) \end{align}\] which is the Taylor’s theorem.