A sequence \(\{p_n\}\) in a metric space \(X\) is said to converge if there is a point \(p\in X\) with the following property: For every \(\epsilon>0\) there is an integer \(N\) such that \(n\ge N\) implies that \(d(p_n,p)<\epsilon\) and we also say \(\{p_n\}\) converges to \(p\) or \(p\) is the limit of \(\{p_n\}\) and we write \[\lim_{x\to \infty}p_n=p\] or \(p_n\to p\). A sequence \(\{p_n\}\) in a metric space \(X\) is said to be a Cauchy sequence if for every \(\epsilon>0\), there is an integer \(N\) such that \(d(p_n,p_m)<\epsilon\) if \(n\ge N\) and \(m\ge N\). The definition of Cauchy sequence does not explicitly involve the limit, which is involved in the definition of convergence. All of the convergent sequences are bounded, because when \(p_n\to p\),\(d(p_n,p)\to 0<M\) for any real number \(M\) and \(p\in X\).
In any metric space \(X\), every convergent sequence is a Cauchy sequence. In \(R^k\), every Cauchy sequence converges. If \(X\) is a compact metric space and if \(\{p_n\}\) is a Cauchy sequence in \(X\), then \(\{p_n\}\) converges to some point in \(X\). A metric space in which every Cauchy sequence converges is said to be complete. All compact metric spaces and all Euclidean spaces are complete.
A sequence \(\{p_{n_i}\}, n_1<n_2<n_3<\cdots\) is called a sebsequence of \(\{p_n\}\) with its items selected from \(\{p_n\}\). Its limit is called a sebsequential limit of \(\{p_n\}\). Let \(E\) be a nonempty subset of a metric space \(X\), and let \(S\) be the set of all real numbers of the form \(d(p,q)\), with \(p\in E\) and \(q\in E\), then the sup of \(S\) is called the diameter of \(E\).
\[\lim_{n\to \infty}\sqrt[n]{n}=1\] because \[n=(\sqrt[n]{n})^n=(1+(\sqrt[n]{n}-1))^n\\ ={n \choose n}1^n(\sqrt[n]{n}-1)^0+{n-1 \choose n}1^{n-1}(\sqrt[n]{n}-1)^1+{n-2 \choose n}1^{n-2}(\sqrt[n]{n}-1)^2+\cdots\\ \ge {n-2 \choose n}1^{n-2}(\sqrt[n]{n}-1)^2=\frac{n(n-1)}{2!}(\sqrt[n]{n}-1)^2\\ \] Then \[(\sqrt[n]{n}-1)\le \sqrt{\frac{2}{n-1}}\] \[\lim_{n\to \infty}\sqrt{\frac{2}{n-1}}=0\] \[\lim_{n\to \infty}(\sqrt[n]{n}-1)=0\]
Given a sequence \(\{a_n\}\) the sum of the sequence \[\sum_{n=1}^{\infty}a_n\] is called a series. The series \(\sum a_n\) converges if and only if for every \(\varepsilon>0\) there is an integer \(N\) such that \[\Bigl|\sum_{k=n}^{m}a_k\Bigr|\leq\varepsilon\quad(m\ge n\ge N)\]
The power series are \[\sum_{n=1}^{\infty}c_nz^n\] 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 \(\sum_{n=1}^{\infty}c_nz^n\), \[\displaystyle\lim_{n\to \infty}\text{sup}\sqrt[n]{|c_nz^n|}=\frac{|z|}{R}\] Then \(\sum_{n=1}^{\infty}c_nz^n\) converges if \(|z|<R\) and diverges if \(|z|>R\).
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 \(\{s_n\}\).