### 数学代写|微积分代写Calculus代写| Cauchy Criterion

statistics-lab™ 为您的留学生涯保驾护航 在代写微积分Calculus方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写微积分Calculus代写方面经验极为丰富，各种代写微积分Calculus相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• Advanced Probability Theory 高等概率论
• Advanced Mathematical Statistics 高等数理统计学
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|微积分代写Calculus代写|Cauchy Criterion

In Theorem 1.1.14, we have given a sufficient condition for the convergence of a sequence $\left(a_{n}\right)$, namely, that if $\left(a_{n}\right)$ satisfies
$$\left|a_{n+2}-a_{n+1}\right| \leq r\left|a_{n+1}-a_{n}\right| \forall n \in \mathbb{N}$$
for some $r$ with $0m$. Thus, $\left|a_{n}-a_{m}\right|$ can be made arbitrarily small for all large enough $n, m \in \mathbb{N}$. Now, we show that any sequence $\left(a_{n}\right)$ such that $\left|a_{n}-a_{m}\right|$ can be made arbitrarily small for all large enough $n, m \in \mathbb{N}$ actually converges. First, let us formally define the requirement on the sequence.

Definition 1.1.13 A a sequence $\left(a_{n}\right)$ is said to be a Cauchy sequence ${ }^{3}$ if for every $\varepsilon>0$, there exists $N \in \mathbb{N}$ such that
$$\left|a_{n}-a_{m}\right|<\varepsilon \quad \forall n, m \geq N .$$
We have already observed in Remark $1.1 .15$ that if $\left(a_{n}\right)$ converges, then it need not satisfy the assumption in Theorem 1.1.14. However, we have the following theorem.
Theorem 1.1.15 Every convergent sequence is a Cauchy sequence.

Proof Suppose $\left(a_{n}\right)$ converges to $a$. Let $\varepsilon>0$ be given. Then we know that there exists $N \in \mathbb{N}$ such that $\left|a_{n}-a\right|<\varepsilon / 2$ for all $n \geq N$. Hence, we have
$$\left|a_{n}-a_{m}\right| \leq\left|a_{n}-a\right|+\left|a-a_{m}\right|<\varepsilon \forall n, m \geq N$$
This completes the proof.
Now, we show that the converse of Theorem 1.1.15 is also true. The idea of the proof is akin to the idea used in the proof of Theorem 1.1.14, namely, we first show that $\left(a_{n}\right)$ is a bounded sequence, so that by Bolzano-Weierstrass theorem (Theorem $1.1 .13),\left(a_{n}\right)$ has a subsequence which converges to some $a$, and then show that $\left(a_{n}\right)$ itself converges to $a$.

Theorem 1.1.16 (Cauchy criterion) Every Cauchy sequence of real numbers converges.

Proof Let $\left(a_{n}\right)$ be a Cauchy sequence. Taking $\varepsilon=1$, there exists $N \in \mathbb{N}$ such that
$$\left|a_{n}-a_{m}\right|<1 \quad \forall n, m \geq N .$$ In particular, for all $n \geq N$, $$\left|a_{n}\right| \leq\left|\left(a_{n}-a_{N}\right)+a_{N}\right| \leq\left|a_{n}-a_{N}\right|+\left|a_{N}\right| \leq 1+\left|a_{N}\right| .$$ Therefore, $$\left|a_{n}\right| \leq \max \left\{\left|a_{1}\right|,\left|a_{2}\right|, \ldots,\left|a_{N}\right|, 1+\left|a_{N}\right|\right\}$$ Thus, $\left(a_{n}\right)$ is a bounded sequence. As already mentioned, by Bolzano-Weierstrass theorem (Theorem $1.1 .13),\left(a_{n}\right)$ has a subsequence $\left(a_{k_{n}}\right)$ which converges to some $a$. Now, let $\varepsilon>0$ be given. Then there exist positive integers $N_{1}, N_{2}$ such that
$$\left|a_{k_{n}}-a\right|<\varepsilon / 2 \quad \forall n \geq N_{1}, \quad\left|a_{n}-a_{k_{n}}\right|<\varepsilon / 2 \quad \forall n \geq N_{2} .$$
Therefore,
$$\left|a_{n}-a\right| \leq\left|a_{n}-a_{k_{n}}\right|+\left|a_{k_{n}}-a\right|<\varepsilon / 2+\varepsilon / 2=\varepsilon$$
for all $n \geq N_{3}:=\max \left{N_{1}, N_{2}\right}$. Thus, $a_{n} \rightarrow a$. This completes the proof.

## 数学代写|微积分代写Calculus代写|Series of Real Numbers

In the last section we have come across sequences whose terms involve some other sequences. For example, we had sequences such as $\left(a_{n}\right)$ with
(i) $a_{n}=\frac{3}{10}+\frac{3}{10^{2}}+\cdots+\frac{3}{10^{n}}$,
(ii) $a_{n}=1+\frac{1}{2}+\cdots+\frac{1}{n}$,
(iii) $a_{n}=1+\frac{1}{2^{2}}+\cdots+\frac{1}{n^{2}}$.
Recall that, the sequence in (i) and (iii) converge whereas the sequence in (ii) diverge. In (i), we have also seen that the sequence converges to $\frac{1}{3}$. Because of this we may represent the number $\frac{1}{3}$ as
$$\frac{1}{3}=\frac{3}{10}+\frac{3}{10^{2}}+\cdots$$
More generally, we may have a sequence $\left(a_{n}\right)$ of real numbers, and we may form a new sequence $\left(s_{n}\right)$ by defining its $n^{\text {th }}$ term as sum of the first $n$ terms of $\left(a_{n}\right)$, that is,
$$s_{n}=a_{1}+\cdots+a_{n} .$$
Then we may enquire whether this new sequence converges or not. In case this sequence $\left(s_{n}\right)$ converge, then we may write its limit as
$$a_{1}+a_{2}+\cdots$$
This expression has a special name!

Definition 1.2.1 A series of real numbers is an expression of the form
$$a_{1}+a_{2}+a_{3}+\ldots$$
or more compactly,
$$\sum_{n=1}^{\infty} a_{n}$$
where $\left(a_{n}\right)$ is a sequence of real numbers. The number $a_{n}$ is called the $n^{\text {th }}$ term of the series and the sum of the first $n$ terms of $\left(a_{n}\right)$, that is,
$$s_{n}:=a_{1}+\cdots+a_{n},$$
is called the $n^{\text {th }}$ partial sum of the series.
Remark 1.2.1 Some authors denote a series as the sequence $\left(a_{n}, s_{n}\right)$, where $s_{n}$ is $n^{\text {th }}$ partial sum of the sequence $\left(a_{n}\right)$.
Example 1.2.1 Following are some examples of series:
(i) $\frac{3}{10}+\frac{3}{10^{2}}+\cdots$,
(ii) $1+\frac{1}{2}+\frac{1}{3}+\cdots$,
(iii) $1+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\cdots$
Note that in (i), (ii), (iii) above the partial sums are
$$\sum_{k=1}^{n} \frac{3}{10^{k}}, \quad \sum_{k=1}^{n} \frac{1}{k}, \quad \sum_{k=1}^{n} \frac{1}{k^{2}}$$
respectively.

## 数学代写|微积分代写Calculus代写|Convergence and Divergence of Series

The following definition is on expected lines (Fig. 1.9):
Definition 1.2.2 A series $\sum_{n=1}^{\infty} a_{n}$ is said to be a convergent series if the corresponding sequence $\left(s_{n}\right)$ of partial sums converges. If $s_{n} \rightarrow s$, then we say that the series $\sum_{n=1}^{\infty} a_{n}$ converges to $s$, and $s$ is called the sum of the series, and we write this fact as
$$s=\sum_{n=1}^{\infty} a_{n}$$

Observe the following: Suppose $a_{n} \geq 0$ for all $n \in \mathbb{N}$. Then the sequence $\left(s_{n}\right)$ of partial sums of the series $\sum_{n=1}^{\infty} a_{n}$ is monotonically increasing. Hence, in this case, either $\left(s_{n}\right)$ converges or $s_{n} \rightarrow \infty$.
Example 1.2.2 Consider the three series given in Example 1.2.1, i.e.,
$$\sum_{n=1}^{\infty} \frac{3}{10^{n}}, \quad \sum_{n=1}^{\infty} \frac{1}{n}, \quad \sum_{n=1}^{\infty} \frac{1}{n^{2}}$$
Note that the partial sums of these series, say $\left(s_{n}^{(1)}\right),\left(s_{n}^{(2)}\right),\left(s_{n}^{(3)}\right)$ with
$$s_{n}^{(1)}:=\sum_{k=1}^{n} \frac{3}{10^{k}}, \quad s_{n}^{(2)}:=\sum_{k=1}^{n} \frac{1}{k}, \quad s_{n}^{(3)}:=\sum_{k=1}^{n} \frac{1}{k^{2}},$$
are the sequences considered in Examples 1.1.5,1.1.19, 1.1.24, respectively, and we have seen that $\left(s_{n}^{(1)}\right)$ and $\left(s_{n}^{(3)}\right)$ converge, whereas $\left(s_{n}^{(2)}\right)$ diverges. Thus, $\sum_{n=1}^{\infty} \frac{3}{10^{n}}$ and $\sum_{n=1}^{\infty} \frac{1}{n^{2}}$ are convergent series, whereas $\sum_{n=1}^{\infty} \frac{1}{n}$ is a divergent series.
Example 1.2.3 Consider the geometric series
$$1+q+q^{2}+\cdots$$
for $q \in \mathbb{R}$. We show that this series converges if and only if $|q|<1$ :
Note that
$$s_{n}=1+q+\cdots+q^{n-1}= \begin{cases}n & \text { if } q=1 \ \left(1-q^{n}\right) /(1-q) & \text { if } q \neq 1\end{cases}$$

## 数学代写|微积分代写Calculus代写|Cauchy Criterion

|一个n+2−一个n+1|≤r|一个n+1−一个n|∀n∈ñ

|一个n−一个米|<e∀n,米≥ñ.

|一个n−一个米|≤|一个n−一个|+|一个−一个米|<e∀n,米≥ñ

|一个n−一个米|<1∀n,米≥ñ.特别是，对于所有n≥ñ,

|一个n|≤|(一个n−一个ñ)+一个ñ|≤|一个n−一个ñ|+|一个ñ|≤1+|一个ñ|.所以，

|一个n|≤最大限度{|一个1|,|一个2|,…,|一个ñ|,1+|一个ñ|}因此，(一个n)是有界序列。如前所述，由 Bolzano-Weierstrass 定理 (Theorem1.1.13),(一个n)有一个子序列(一个ķn)收敛到一些一个. 现在，让e>0被给予。那么存在正整数ñ1,ñ2这样

|一个ķn−一个|<e/2∀n≥ñ1,|一个n−一个ķn|<e/2∀n≥ñ2.

|一个n−一个|≤|一个n−一个ķn|+|一个ķn−一个|<e/2+e/2=e

## 数学代写|微积分代写Calculus代写|Series of Real Numbers

（一）一个n=310+3102+⋯+310n,
(ii)一个n=1+12+⋯+1n,
(iii)一个n=1+122+⋯+1n2.

13=310+3102+⋯

sn=一个1+⋯+一个n.

∑n=1∞一个n

sn:=一个1+⋯+一个n,

(i)310+3102+⋯,
(ii)1+12+13+⋯,
(iii)1+122+132+⋯

∑ķ=1n310ķ,∑ķ=1n1ķ,∑ķ=1n1ķ2

## 数学代写|微积分代写Calculus代写|Convergence and Divergence of Series

s=∑n=1∞一个n

∑n=1∞310n,∑n=1∞1n,∑n=1∞1n2

sn(1):=∑ķ=1n310ķ,sn(2):=∑ķ=1n1ķ,sn(3):=∑ķ=1n1ķ2,

1+q+q2+⋯

sn=1+q+⋯+qn−1={n 如果 q=1 (1−qn)/(1−q) 如果 q≠1

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。