Posted on Categories:Measure Theory and Fourier Analysis, 数学代写, 测度论和傅里叶分析

# 数学代写|测度论和傅里叶分析代写Measure Theory and Fourier Analysis代考|MATH4069 Consider the sum

avatest™

## avatest™帮您通过考试

avatest™的各个学科专家已帮了学生顺利通过达上千场考试。我们保证您快速准时完成各时长和类型的考试，包括in class、take home、online、proctor。写手整理各样的资源来或按照您学校的资料教您，创造模拟试题，提供所有的问题例子，以保证您在真实考试中取得的通过率是85%以上。如果您有即将到来的每周、季考、期中或期末考试，我们都能帮助您！

•最快12小时交付

•200+ 英语母语导师

•70分以下全额退款

## 数学代写|测度论和傅里叶分析代写Measure Theory and Fourier Analysis代考|Consider the sum

Consider the sum $\sum_{n=1}^{\infty} \frac{1}{(n+2)(n+1)}$.
a. Use an appropriate convergence test to show that this series converges.
Since the tail of the series determines the convergence, we note that for large $n$
$$\frac{1}{(n+2)(n+1)} \sim \frac{1}{n^2} .$$
So, we can compare the original series to $\sum_{n=1}^{\infty} \frac{1}{n^2}$. Since the latter series converges, so does this one by the Limit Comparison Test.
One can also use the following
$$\sum_{n=1}^{\infty} \frac{1}{(n+2)(n+1)}<\sum_{n=1}^{\infty} \frac{1}{n^2}<\infty .$$
Therefore, the series converges by the Comparison Test as well.
b. Verify that
$$\sum_{n=1}^{\infty} \frac{1}{(n+2)(n+1)}=\sum_{n=1}^{\infty}\left(\frac{n+1}{n+2}-\frac{n}{n+1}\right) .$$
One just adds the terms to verify the sum.
\begin{aligned} \frac{n+1}{n+2}-\frac{n}{n+1} &=\frac{(n+1)^2-n(n+2)}{(n+2)(n+1)} \ &=\frac{1}{(n+2)(n+1)} . \end{aligned}
Note that partial fractions does not give this representation, but instead gives
$$\sum_{n=1}^{\infty} \frac{1}{(n+2)(n+1)}=\sum_{n=1}^{\infty}\left(\frac{1}{n+1}-\frac{1}{n+2}\right) .$$

## 数学代写|测度论和傅里叶分析代写Measure Theory and Fourier Analysis代考|Recall that the alternating harmonic series converges conditionally

a. From the Taylor series expansion for $f(x)=\ln (1+x)$, inserting $x=1$ gives the alternating harmonic series. What is the sum of the alternating harmonic series?
The Taylor series expansion for $f(x)=\ln (1+x)$ is given by
$$\ln (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\ldots=\sum_{n=1}^{\infty}(-1)^{n+1} \frac{x^n}{n}$$
Inserting $x=1$ gives the sum of the alternating harmonic series
$$\ln 2=1-\frac{1}{2}+\frac{1}{3}-\ldots=\sum_{n=1}^{\infty}(-1)^{n+1} \frac{1}{n} .$$
b Because the alternating harmonic series does not converge absolutely, a rearrangement of the terms in the series will result in series whose sums vary. One such rearrangement in alternating $p$ positive terms and $n$ negative terms leads to the following sum ${ }^1$ :
\begin{aligned} \frac{1}{2} \ln \frac{4 p}{n}=\underbrace{\left(1+\frac{1}{3}+\cdots+\frac{1}{2 p-1}\right)}{p \text { terms }}-\underbrace{\left(\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2 n}\right)}{n \text { terms }} \ &+\underbrace{\left(\frac{1}{2 p+1}+\cdots+\frac{1}{4 p-1}\right)}{p \text { terms }}-\underbrace{\left(\frac{1}{2 n+2}+\cdots+\frac{1}{4 n}\right)}{n \text { terms }}+\cdots . \end{aligned}
Find rearrangements of the alternating harmonic series to give the following sums; that is, determine $p$ and $n$ for the given expression and write down the above series explicitly; that is, determine $p$ and $n$ leading to the following sums.

## 数学代写测度论和傅里叶分析代写Measure Theory and Fourier Analysis代考|Consider the sum

$$\frac{1}{(n+2)(n+1)} \sim \frac{1}{n^2} .$$

$$\sum_{n=1}^{\infty} \frac{1}{(n+2)(n+1)}<\sum_{n=1}^{\infty} \frac{1}{n^2}<\infty .$$

$$\sum_{n=1}^{\infty} \frac{1}{(n+2)(n+1)}=\sum_{n=1}^{\infty}\left(\frac{n+1}{n+2}-\frac{n}{n+1}\right) .$$
$$\frac{n+1}{n+2}-\frac{n}{n+1}=\frac{(n+1)^2-n(n+2)}{(n+2)(n+1)} \quad=\frac{1}{(n+2)(n+1)} .$$

$$\sum_{n=1}^{\infty} \frac{1}{(n+2)(n+1)}=\sum_{n=1}^{\infty}\left(\frac{1}{n+1}-\frac{1}{n+2}\right) .$$

## 数学代写测度论和傅里叶分析代写Measure Theory and Fourier Analysis代考|Recall that the alternating harmonic series converges conditionally

$$\ln (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\ldots=\sum_{n=1}^{\infty}(-1)^{n+1} \frac{x^n}{n}$$
$$\ln 2=1-\frac{1}{2}+\frac{1}{3}-\ldots=\sum_{n=1}^{\infty}(-1)^{n+1} \frac{1}{n} .$$

$$\frac{1}{2} \ln \frac{4 p}{n}=\underbrace{\left(1+\frac{1}{3}+\cdots+\frac{1}{2 p-1}\right)} p \text { terms }-\underbrace{\left(\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2 n}\right)} n \text { terms }+\underbrace{\left(\frac{1}{2 p+1}+\cdots+\frac{1}{4 p-1}\right)} p \text { terms }-\underbrace{\left(\frac{1}{2 n+2}+\cdots+\frac{1}{4 n}\right)} n$$

## 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。