Sequences and Series

sequences and series数学讲义概述及范文模板1. 引言1.1 概述在数学领域中，序列和级数是重要的概念，它们在许多实际问题的建模和解决中起着至关重要的作用。

序列由一系列按照特定规律排列的数构成，而级数则是将序列中的数相加得到的结果。

这些概念被广泛应用于计算机科学、物理学、经济学等多个领域。

本篇文章将深入介绍序列和级数的基本定义、性质以及相关定理。

通过阐述这些重要概念，读者将能够更好地理解它们在实际问题中的运用，并且掌握一些常见的求解方法。

1.2 文章结构本文分为引言、正文和结论三个部分。

引言部分将对文章整体进行简要介绍，包括序列和级数概念的概述、本文目的以及文章结构。

正文部分将详细阐述序列和级数的基本概念、性质以及求解方法。

每个章节将围绕一个特定主题展开，结合范例和推导过程深入讲解相关知识点。

结论部分将对全文进行总结，并提供一些进一步学习的建议和参考资料。

1.3 目的本文的目的在于引导读者全面了解序列和级数的概念、定义和性质，并掌握一些常用的解题方法。

通过对本文内容的学习，读者将能够应对实际问题中涉及序列和级数的计算及分析，并进一步拓展数学思维和推理能力。

在阅读本文之前，读者需要具备一定的数学基础知识，包括初等代数、函数以及各种基本运算规则等。

这些基础将有助于更好地理解和应用本文中所涉及到的概念和定理。

总之，希望本文能为读者提供一个扎实而全面的关于序列和级数的讲义，在深入研究该领域或解决实际问题时起到指导作用。

接下来我们将进入正文部分，详细介绍序列和级数相关知识。

2. 正文在数学中，序列和级数是重要的概念。

序列是一组按照特定顺序排列的数，而级数则是将序列中的所有项进行求和得到的结果。

本文将详细介绍序列和级数的性质、定义以及其重要应用领域。

首先，我们来看序列。

一个序列可以由各种规则生成，例如公式、递推关系或某种算法。

每个序列包含一系列有限或无限个数字，并按照特定的次序排列。

其中，有限序列是指元素数量有限的序列，而无限序列则是指元素数量无限的序列。

algebra2 知识点总结Linear Equations and FunctionsOne of the fundamental concepts in Algebra 2 is linear equations and functions. A linear equation is an equation of the form y = mx + b, where m is the slope and b is the y-intercept. Students learn how to graph linear equations, find the slope and y-intercept, and solve systems of linear equations using various methods such as substitution, elimination, and graphing.Functions are a core concept in Algebra 2, and students study various types of functions such as linear, quadratic, exponential, and logarithmic functions. They learn how to analyze the behavior of functions, find their domain and range, and determine whether a function is even, odd, or periodic. Students also explore transformations of functions, such as shifts, stretches, and reflections, and how they affect the graph of a function.Inequalities and Absolute Value EquationsIn Algebra 2, students also study inequalities and absolute value equations. They learn how to solve and graph linear inequalities, quadratic inequalities, rational inequalities, and absolute value inequalities. They also explore the concept of compound inequalities and how to solve systems of inequalities.Absolute value equations are another important topic in Algebra 2. Students learn how to solve and graph absolute value equations, as well as inequalities involving absolute value expressions. They also study the properties of absolute value functions and their applications in real-life scenarios.Polynomials and Polynomial FunctionsPolynomials are algebraic expressions that consist of variables and coefficients, combined using addition, subtraction, and multiplication. In Algebra 2, students learn how to add, subtract, multiply, and divide polynomials, as well as factor and solve polynomial equations. They also study the properties of polynomial functions, such as end behavior, zeros, and turning points.Students delve into advanced topics such as polynomial long division, synthetic division, the remainder theorem, and the factor theorem. They also explore the relationship between polynomial functions and their graphs, and how to use this information to solve real-world problems.Rational Expressions and Rational FunctionsRational expressions are fractions that contain polynomials in the numerator and denominator. In Algebra 2, students learn how to simplify, multiply, divide, add, and subtract rational expressions, as well as solve rational equations. They also study the properties of rational functions, such as asymptotes, intercepts, and end behavior.Students explore the relationship between rational functions and their graphs, and how to use this information to analyze and solve real-world problems. They also study advanced topics such as partial fraction decomposition, complex fractions, and applications of rational functions in areas such as economics, physics, and engineering.Exponential and Logarithmic FunctionsExponential and logarithmic functions are essential in Algebra 2, and students learn how to solve exponential and logarithmic equations, as well as graph exponential and logarithmic functions. They study the properties of exponential and logarithmic functions, such as growth and decay, domain and range, and asymptotic behavior.Students also explore the relationship between exponential and logarithmic functions, and how to use this information to solve real-world problems. They study applications of exponential and logarithmic functions in areas such as finance, population growth, radioactive decay, and pH levels.Sequences and SeriesSequences and series are important topics in Algebra 2, and students learn how to find the nth term of a sequence, as well as the sum of a finite or infinite series. They study arithmetic sequences and series, geometric sequences and series, and other types of sequences such as Fibonacci and recursive sequences.Students explore the properties of sequences and series, such as common difference, common ratio, and convergence. They also study applications of sequences and series in areas such as finance, physics, and computer science.Complex NumbersComplex numbers are numbers of the form a + bi, where a and b are real numbers, and i is the imaginary unit (√-1). In Algebra 2, students learn how to perform operations with complex numbers, such as addition, subtraction, multiplication, division, and simplification. They also study the properties of complex numbers, such as the conjugate, modulus, and argument.Students explore the relationship between complex numbers and their graphs on the complex plane, and how to use this information to solve equations involving complex numbers. They also study applications of complex numbers in areas such as electrical engineering, signal processing, and quantum mechanics.Matrices and DeterminantsMatrices are arrays of numbers arranged in rows and columns, and they are used to represent and solve systems of linear equations. In Algebra 2, students learn how to add, subtract, multiply, and invert matrices, as well as find the determinant of a matrix. They also study the properties of matrices, such as the identity matrix, transpose, and rank.Students explore the relationship between matrices, determinants, and systems of linear equations, and how to use this information to solve real-world problems. They also study applications of matrices in areas such as computer graphics, cryptography, and economics.Conic SectionsConic sections are curves obtained by intersecting a cone with a plane, and they include the circle, ellipse, parabola, and hyperbola. In Algebra 2, students learn how to graph and analyze conic sections, as well as find their equations given certain properties.Students study the properties of conic sections, such as the focus, directrix, eccentricity, and asymptotes. They also explore the relationship between conic sections and their equations, and how to use this information to solve real-world problems. They study applications of conic sections in areas such as astronomy, engineering, and architecture.ConclusionAlgebra 2 is a challenging but rewarding branch of mathematics that builds upon the concepts learned in Algebra 1. In this article, we have provided a comprehensive summary of the key topics in Algebra 2, including linear equations and functions, inequalities and absolute value equations, polynomials and polynomial functions, rational expressions and rational functions, exponential and logarithmic functions, sequences and series, complex numbers, matrices and determinants, and conic sections.By mastering these topics, students will develop a deeper understanding of algebraic concepts and techniques, as well as their applications in various fields such as science, engineering, economics, and finance. Algebra 2 is an essential foundation for further study in mathematics and related disciplines, and it provides students with the analytical and problem-solving skills necessary for success in the modern world.。

Core Mathematics1（AS/A2）——核心数学11.Algebra and functions——代数和函数2.Quadratic functions——二次函数3.Equations and inequalities——等式和不等式4.Sketching curves——画图（草图）5.Coordinate geometry in the （x，y）plane——平面坐标系中的坐标几何6.Sequences and series——数列7.Differentiation——微分8.Integration——积分Core Mathematics2（AS/A2）——核心数学21.Algebra and functions——代数和函数2.The sine and cosine rule——正弦和余弦定理3.Exponentials and logarithm——指数和对数4.Coordinate geometry in the （x，y）plane——平面坐标系中的坐标几何5.The binomial expansion——二项展开式6.Radian measure and its application——弧度制及其应用7.Geometric sequences and series——等比数列8.Graphs of trigonometric functions——三角函数的图形9.Differentiation——微分10.Trigonometric identities and simple equations——三角恒等式和简单的三角等式11.Integration——积分Core Mathematics3（AS/A2）——核心数学31.Algebra fractions——分式代数2.Functions——函数3.The exponential and log functions——指数函数和对数函数4.Numerical method——数值法5.Transforming graph of functions——函数的图形变换6.Trigonometry——三角7.Further trigonometric and their applications——高级三角恒等式及其应用8.Differentiation——微分Core Mathematics4（AS/A2）——核心数学41.Partial fractions——部分分式2.Coordinate geometry in the （x，y）plane——平面坐标系中的坐标几何3.The binomial expansion——二项展开式4.Differentiation——微分5.Vectors——向量6.Integration——积分A-Level：核心数学Core Maths，力学数学，统计数学，决策数学Core Mathematics1（AS/A2）——核心数学11.Algebra and functions——代数和函数2.Quadratic functions——二次函数3.Equations and inequalities——等式和不等式4.Sketching curves——画图（草图）5.Coordinate geometry in the （x，y）plane——平面坐标系中的坐标几何6.Sequences and series——数列7.Differentiation——微分8.Integration——积分每章内容：Core Mathematics2（AS/A2）——核心数学21.Algebra and functions——代数和函数2.The sine and cosine rule——正弦和余弦定理3.Exponentials and logarithm——指数和对数4.Coordinate geometry in the （x，y）plane——平面坐标系中的坐标几何5.The binomial expansion——二项展开式6.Radian measure and its application——弧度制及其应用7.Geometric sequences and series——等比数列8.Graphs of trigonometric functions——三角函数的图形9.Differentiation——微分10.Trigonometric identities and simple equations——三角恒等式和简单的三角等式11.Integration——积分每章内容：1.Algebra fractions——分式代数2.Functions——函数3.The exponential and log functions——指数函数和对数函数4.Numerical method——数值法5.Transforming graph of functions——函数的图形变换6.Trigonometry——三角7.Further trigonometric and their applications——高级三角恒等式及其应用8.Differentiation——微分每章内容：1.Partial fractions——部分分式2.Coordinate geometry in the （x，y）plane——平面坐标系中的坐标几何3.The binomial expansion——二项展开式4.Differentiation——微分5.Vectors——向量6.Integration——积分每章内容：。

Chapter 10 Sequences and Series序列和级数【Vocabulary · 词汇梳理】【导图】A. Sequences of Real Numbers实数序列An infinite sequence is a function whose domain is the set of positive integers, and is often denoted simply by a n hhe siquence defined for example,by a n=1nis the set of numbers1,12,13,…,1n,…hhe elements in this set are called the terms of the sequence, and the n th or generalterm of this sequence is 1n·A sequence a n converges to a finite number L if limn→∞a n=L If a n does not have a (finite) limit, we say the sequence is divergentExample 1Does the sequence a n=1nconverge or diverge?Solution:lim n→∞1n=0; hence the sequence converges to 0Example 2Does the sequence a n=3n4+54n4−7n2+9converge or diverge? Solution:lim n→∞3n4+54n4−7n2+9=34; hence the sequence converges to 34Example 3Does the sequence a n=1+(−1)nnconverge or diverge? Solution:lim n→∞1+(−1)nn=1, hence the sequence converges to 1 Note that the terms in the sequence0,32,23,54,45,76,… are alternately smaller and larger than 1 We say this sequence converges to 1 byoscillationExample 4Does the sequence a n =n 2−1nconverge or diverge?Solution:limn→∞n 2−1n=∞, the sequence diverges (to infinity)Example 5Does the sequence a n =sin n converge or diverge? Solution:Because lim n→∞sin n does not exist, the sequence diverges However, note that it does notdiverge to infinityExample 6Does the sequence a n =(−1)n+1 converge or diverge? Solution:Because lim n→∞(−1)n+1 does not exist, the sequence diverges Note that the sequence1,−1,1,−1… diverges because it oscillatesB. INFINITE SERIES 无穷级数B1. Definitions 定义If a n is a sequence of real number, then an infinite series is an expression of the form∑a k ∞k=1=a 1+a 2+a 3+⋯+a n +⋯ (1)hhe elements in the sum are called terms ; a n is the n th or general term of the series Example 7A series of the form ∑1k p∞k=1is called a p-series (p 级数)hhe p-series for p =2 is ∑1k 2∞k=1=112+122+132+⋯+1n 2+⋯Example 8hhe p-series for p =1 is called the harmonic series (调和级数):∑1k ∞k=1=11+12+13+⋯+1n+⋯Example 9A geometric series(几何级数) has a first term, a, and common ration of terms, r∑ar k−1∞k=1=a+ar+ar2+ar3+⋯+ar n−1+⋯If there is a finite number S such thatlim n→∞∑a k∞k=1=Sthen we say that infinite series is convergent, or converges to S, or has the sum S, and we write in this case,∑a k∞k=1=SWhen there is no source of confusion, the infinite series (1)may be indicated simply by∑a k or ∑a nExample 10Show that the geometric series 1+12+14+⋯+12n+⋯converges to 2Solution:Let S represent the sum of the series, then:S=limn→∞(1+1+1+⋯+1n+⋯)1 2S=limn→∞(12+14+⋯+12n+⋯)Subteaction yields1S=limn→∞(1−1n+1)Hence, S=2 Example 11Show that the harmonic series1+12+13+14+⋯+1n+⋯divergesSolution:hhe terms in the series can be grouped as follows:1+12+(13+14)+(15+16+17+18)+(19+110+⋯+116)+(117+⋯+132)+⋯hhis sum clearly exceeds1+12+2(14)+4(18)+8(116)+16(132)+⋯1+12+12+12++⋯Since that sum is not bounded, it follows that ∑1n diverges to ∞B2. Theorems About Convergence or Divergence of Infinite Series 无穷级数的收敛和发散定理Theorem 2a. If ∑a k converges, the lim n→∞a n =0hhis provides a convenient and useful test for divergence, since it is equivalent to the statement: If a n does not approach zero, then the series ∑a k diverges Note, however, particularly that the converse of Theorem 2a is not true hhe condition that a n approach zero is necessary but not sufficient (必要非充分条件) for the convergence of the serieshhe harmonic series ∑1n is an excellent example of a series whose n th term goes to zero butthat diverges (see Example 11 above) hhe series ∑nn+1diverges because lim n→∞a n =1, not zero;the series ∑n n 2+1does not converge (as will be shown shortly) even though lim n→∞a n =0Theorem 2b. A finite number of terms may be added to or deleted from a series without affecting its convergence or divergence; thus∑a k ∞k=1 and ∑a k ∞k=m(where m is any positive integer) both converge or both diverge (Note that the sums most likely will differ )Theorem 2c. hhe terms of a series may be multiplied by a nonzero constant without affecting the convergence or divergence; thus∑a k ∞k=1 and ∑ca k ∞k=1Both converge or both diverge (Again, the sums will usually differ ) Theorem 2d. If ∑a n and ∑b n both converge, so does ∑(a n +b n )Theorem 2e. If the terms of a convergent series are regrouped, the new series convergesB3. Tests for Convergence of Infinite Series 无穷级数的收敛判别法 THE nth TERM TEST 尾项判别法If lim n→∞a n ≠0, then ∑a n divergesNOTE : When working with series, it’s a good idea to start by checking the n th herm hest If the terms don’t approach 0, the series cannot converge hhis is often the quickest and easiest way to identify a divergent series(Because this is the contrapositive of hheorem 2a, it’s always true But beware of the converse! Seeing that the terms do approach 0 does not guarantee that the series must converge. It just means that you need to try other tests )Does ∑n 2n+1converge or diverge?Solution: Since limn→∞n 2n+1=12≠0, the series ∑n2n+1 diverges by the n th herm hestTHE GEOMETRIC SERIES TEST 几何级数判别法 A geometric series ∑ar n converges if and only if |r |<1 If |r |<1, the sum of the series isa 1−rhhe series cannot converge unless it passes the n th herm hest; lim n→∞ar n =0 only if |r |<1As noted earlier, this is a necessary condition for convergence, but may not be sufficient We now examine the sum using the same technique we employed in Example 10S =lim n→∞(a +ar +ar 2+ar 3+⋯+ar n )rS =lim n→∞(ar +ar 2+ar 3+⋯+ar n +ar n+1)(1−r )S =lim n→∞(a −ar n+1)=a −lim n→∞ar n+1 (and remember:|r |<1)=a S =a 1−rExample 13Does 0.3+0.03+0.003+⋯ converge or diverge? Solution:hhe series 0.3+0.03+0.003+⋯ is geometric with a =0.3 and r =0.1 Since |r |<1, the series converges, and its sum isS =a 1−r =0.31−0.1=0.30.9=13Note : 13=0.3333…, which is the given seriesB4. Tests for Convergence of Nonnegative series 正项级数的收敛判别法 hhe series ∑a n is called a nonnegative series if a n ≥0 for all nTHE INTEGRAL TEST 积分判别法Let ∑a n be a nonnegative series If f(x) is a continuous, positive, decreasing function and f (n )=a n , then ∑a n converges if and only if the improper integral ∫f (x )dx ∞1 convergesDoes ∑n n 2+1converge?Solution:hhe associated improper integral is∫xdx x 2+1∞1=lim b→∞12ln (x 2+1)|1b=∞hhe improper integral and the infinite series both divergeExample 15hest the series ∑ne n for convergenceSolution:∫x e x dx ∞1=lim b→+∞∫xe −x dx b 1=lim b→+∞−e −x (1+x )|1b =−lim b→+∞(1+b e b −2e )=2eBy an application of L’Hôpital’s Rule hhus ∑n e nconvergesTHE p -SERIES TEST p -级数判别法A p -series ∑1n p ∞n=1 converges if p >1, but diverges if p ≤1hhis follows immediately from the Integral hest and the behavior of improper integrals of the form ∫1x pdx ∞1Example 16Does the series 1+123+133+⋯+1n 3+⋯ converge or diverge? Solution: hhe series 1+123+133+⋯+1n 3+⋯ is a p -series with p =3, hence the series converges bythe p -Series hestExample 17Does the series √n converge or diverge?Solution: ndiverges, because it is a p -series with p =12THE COMPARISON TEST 比较判别法We compare the general term of ∑a n , the nonnegative series we are investigating, with the general term of a series, ∑u n , known to converge or diverge(1)(1) If ∑u n converges and a n≤u n, then ∑a n converges(2)(2) If ∑u n diverges and a n≥u n, the ∑a n divergesAny known series can be used for comparison Particularly useful are p-series, which converge if p>1but diverge if p≤1, and geometric series, which converge if |r|<1but diverge if |r|≥1Example 18Does the series ∑11+n4converge or diverge?Solution:Since 11+n4<1n4and the p-series ∑1n4converges, ∑11+n4converges by the Comparison hestExample 19Does the series√2√5√8+⋯+√3n−1⋯converge or diverge?Solution:√2√5√8+⋯√3n−1⋯diverges, since√3n−1>√3n=1√3∙n12the latter is the general term of the divegent p-series ∑cn p , where c=3and p=12Remember in using the Comparison hest that you may either discard a fìnite number of terms or multiply each term by a nonzero constant without affecting the convergence of the seríes you are testíngExample 20Does the series ∑1n n =1+122+133+⋯+1n n+⋯convergeSolution:For n>2,1n n <12nand ∑12nis a convergent geometric series with r=12THE LIMIT COMPARISON TEST 极限比较判别法Let ∑a n be a nonnegative series that we are investigating Given ∑b n, a nonnegative series known to be convergent or divergent:(1)If limn→∞a nb n=L, where 0<L<∞, then ∑a n and ∑b n both converge or diverge(2)If limn→∞a nb n=0, and ∑b n converges, then ∑a n converges(3)If limn→∞a nb n=∞, and ∑b n diverges, then ∑a n divergesAny known series can be used for comparison Particularly useful are p-series, which converge if p>1but diverge if p≤1,and geometric series, which converge if |r|<1but diverge if |r|≥1hhis test is useful when the direct comparisons required by the Comparison hest are difficult to establish or when the behavior of ∑a n is like that of ∑b n, but the comparison of the individual terms is in the wrong direction necessary for the Comparison hest to be conclusiveExample 21Does ∑12n+1convergeSolution:hhis series seems to be related to the divergent harmonic series, but 12n+1<1n, so thecomparison fails However, the Limit Comparison hest yields:lim n→∞12n+11n=limn→∞n=1Since ∑1n diverges, ∑12n+1also diverges by the Limit Comparison hestTHE RATIO TEST 比值判别法Let ∑a n be a nonnegative series, and let limn→∞a n+1a n=L, if it exists hhen ∑a n converges ifL<1and diverges if L>1If L=1, this test is inconclusive; apply one of the other testsNOTE: It is good practice, when using the ratio test, to first write limn→∞|a n+1a n|; then, if it isknown that the ratio is always nonnegative, you may rewrite the limit without the absolute value However, when using the ratio test on a power series, you must retain the absolute value throughout the limit process because it could be possible that x<0Example 22Does ∑1n!converge or diverge?Solution:lim n→∞a n+1a n=limn→∞1(n+1)!1n!=limn→∞n!(n+1)!=limn→∞1n+1=0hherefore this series converges by the Ratio hest Example 23Does ∑n nn!converge or diverge?Solution:a n+1n =(n +1)n+1()∙n!n =(n +1)n n =(n +1)nlim n→∞(n +1n )n =lim n→∞(1+1n)n =e Since e >1,∑n n n!diverges by the Ratio hestExample 24If the Ratio het is applied to any p -series, ∑1n p , thena n+1a n =1(n +1)p 1n p =(n n +1)plim n→∞(n n +1)p=1 for all p But if p >1 then ∑1n p converges, while if p ≤1 then ∑1n p diverges hhis illustrates the failure of the Ratio hest to resolve the question of convergence when the limit of the ratio is 1THE ROOT TEST 根值判别法Let lim n→∞√a n n =L , if it exists hhen ∑a n converges if L <1 and diverges if L >1If L =1 this test is inconclusive; try one of the other testshhe decision rule for this test is the same as that for the Ratio hestNOhE: hhe Root hest is not specifically tested on the AP Calculus ExamExample 25 hhe series ∑(n 2n+1)nconverges by the Root hest, sincelim n→∞√(n )n n=lim n→∞n =1B5. Alternating Series and Absolute Convergence 交错级数和绝对收敛Any test that can be applied to a nonnegative series can be used for a series all of whose terms are negative We consider here only one type of series with mixed signs, the so-called alternating series hhis has the form:∑(−1)k+1a k ∞k=1=a 1−a 2+a 3−a 4+⋯+(−1)k+1a k +⋯Where a k >0hhe series1−12+13−14+⋯+(−1)n+1∙1n+⋯is the alternating harmonic seriesTHE ALTERNATING SERIES TEST 交错级数判别法An alternating series converges if:(1)a n+1<a n for all n, and(2)limn→∞a n=0Example 26Does the series ∑(−1)n+1nconverge or diverge? Solution:hhe alternating harmonic series ∑(−1)n+1nconverges, since(1)1n+1<1nfor all n and(2)limn→∞1n=0Example 27Does the series 12−23+34−⋯converge or diverge?Solution:hhe series 12−23+34−⋯diverges, since we see that limn→∞nn+1=1,not 0(By the n thhermhest, if a n does not approach 0, then ∑a n does not converge )ABSOLUTE CONVERGENCEAND CONDITIONAL CONVERGENCE绝对收敛和条件收敛A series with mixed signs is said to converge absolutely (or to be absolutely convergent) if the series obtained by taking the absolute values of its terms converges; that is, ∑a n converges absolutely if ∑|a n|=|a1|+|a2|+⋯+|a n|+⋯convergesA series that converges but not absolutely is said to converge conditionally(or to be conditionally convergent) hhe alternating harmonic series converges conditionally since it converges, but does not converge absolutely (hhe harmonic series diverges )When asked to determine whether an alternating series is absolutely convergent, conditionally convergent, or divergent, it is often advisable to first consider the series of absolute values Check first for divergence, using the n th herm hest If that test shows that the series may converge, investigate further, using the tests for nonnegative series If you find that the series of absolute valuesconverges, then the alternating series is absolutely convergent If, however, you find that the series of absolute values diverges, then you'll need to use the Alternating Series hest to see whether the series is conditionally convergentExample 28Determine whether ∑(−1)n n 2n 2+9converges absolutely, converges conditionally, or divergesSolution: We see that lim n→∞n 2n 2+9=1, not 0, so by the n th herm hest the series ∑(−1)n n 2n 2+9is divegentExample 29 Determine whether ∑sinnπ3n2 converges absolutely, converges conditionally, or divergesSolution:Note that, since |sin nπ3|≤1,limn→∞sinnπ3n 2=0 ; the series passes the n th herm hest Also,|sinnπ3n|≤1n for all nBut 1n 2is the general term of a convergent p -series (p =2), so by the Comparison hest thnonnegative series converges, and therefore the alternation series converges absolutelyExample 30Determine whether n+1√n+13converges absolutely, converges conditionally, or divergesSolution: √n+13is a p -series with p =13, so the nonnegative series divergesWe see that ()3<n+13and limn+13=0, so the alternating series converges; hencen+1√n+13is conditionally convergentAPPROXIMATING THE LIMIT OF AN ALTERNATING SERIES 交错级数的近似极限Evaluating the sum of the first n terms of an alternation series, given by ∑(−1)k+1a k n k=1, yields an approximation of the limit, L hhe error (the difference between the approximation and the true limit) is called the remainder after n terms and is denoted by R n When an alternating series is first shown to pass the Alternating Series hest ，it’s easy to place an upper bound on this remainder Because the terms alternate in sign and become progressively smaller in magnitude, an alternating series converges on its limit by oscillation, as shown in Figure 10-1Figure 10- 1Because carrying out the approximation one more term would once more carry us beyond L , we see that the error is always less than that next term Since |R n |<a n+1 the alternating series error bound for an alternating series is the first term omitted or droppedExample 31hhe series∑(−1)k+1k∞k=1 passes the Alternating Series hest, hence its sum differs from the sum(1−12+13−14+15−16)by less than 17, which is the error boundExample 32Use the alternating series error bound to determine how many terms must be summed to approximate to three decimal places the value of 1−14+19−116+⋯+(−1)n+1n 2+⋯?Solution: Since1(n+1)2<1n 2 and lim n→∞1n 2=0 , the series converges by the Altenating Series hest;therefore after summing a number of terms the remainder (alternating series error bound) will beless than the first omitted termWe seek n such that R n =1(n+1)2<0.001 hhus n must satisfy (n +1)2>1000, or n >30.623 hherefore 31 terms are needed for the desired accuracyC. POWER SERIES 幂级数C1. Definitions; Convergence 定义; 收敛 An expression of the form∑a k x k ∞k=0=a 0+a 1x +a 2x 2+⋯+a n x n +⋯, (1)Where the a’s are constants, is called a power series in x ; and∑a k (x −a)k ∞k=0=a 0+a 1(x −a)+a 2(x −a )2+⋯+a n (x −a )n +⋯, (2)is called a power series in (x −a)If in (1) or (2) x is replaced by a specific real number, then the power series becomes a series of constants that either converges or diverges Note that series (1) converges if x =0 and series (2) converges if x =aRADIUS AND INTERV AL OF CONVERGENCE 收敛半径和收敛区间If power series (1) converges when |x |<r and diverges when |x |>r , then r is called the radius of convergence Similarly, r is the radius of convergence of power series (2) if (2) converges when |x −a |<r and diverges when |x −a |>rhhe set of all values of x for which a power series converges is called its interval of convergence ho find the interval of convergence, first determine the radius of convergence by applying the Ratio hest to the series of absolute values hhen check each endpoint to determine whether the series converges or diverges thereExample 33Find all x for which the following series converges:1+x +x 2+⋯+x n +⋯ (3)Solution:By the Ratio hest, the series converges iflim n→∞|u n+1u n |=lim n→∞|x n+1x n |=lim n→∞|x |=|x |<1 hhus, the radius of convergence is 1 hhe endpoints must be tested separately since the Ratiohest fails when the limit equals 1 When x =1, (3) becomes 1+1+1+⋯ and diverges; when x =−1, (3) becomes 1−1+1−1+⋯ and diverges hhus the interval of convergence is −1<x <1Example 34For what x does ∑(−1)n−1x n−1n+1∞n=1 converge?Solution:lim n→∞|u n+1u n |=lim n→∞|x n n +2∙n +1x n−1|=lim n→∞|x |=|x |<1 hhe radius of convergence is 1 When x =1 , we have 12−13+14−15+⋯ , an alternating convergent series; when x =−1 , the series is 12+13+14+⋯ , which diverges hhus, the series converges if −1<x ≤1Example 35For what values of x does ∑x n n!∞n=1converge?Solution:lim n→∞|u n+1u n |=lim n→∞|x n+1(n +1)!∙n!x n |=lim n→∞|x |n +1=0 which is always less than 1 hhus the series converges for all xExample 36Find all x for which the following series converges:1+x −21+(x −2)22+⋯+(x −2)n−1n−1+⋯ (4)Solution:lim n→∞|u n+1u n |=lim n→∞|(x −2)n 2n ∙2n−1(x −2)n−1|=lim n→∞|x −2|2=|x −2|2which is less than 1 if |x −2|<2, that is, if 0<x <4 Series (4) converges on this intervaland diverges if |x −2|>2, that is, if x <0 or x >4 When x =0, (4) is 1−1+1−1+⋯ and diverges When x =4, (4) is 1+1+1+⋯ and diverges hhus, the interval of convergence is 0<x <4Example 37Find all x for which the series ∑n!x n∞n=1 converges Solution:∑n!x n ∞n=1 converges only at x =0, sincelim n→∞u n+1u n =lim n→∞(n +1)x =∞ unless x =0C2. Functions Defined by Power Series 幂级数定义的函数 Let the function f be defined byf (x )=∑a k (x −a)k ∞k=0=a 0+a 1(x −a )+⋯+a n (x −a )n +⋯ (1)its domain is the interval of convergence of the seriesFunctions defined by power series behave very much like polynomials, as indicated by the following properties:PROPERTY 2a. hhe function defined by (1) is continuous for each x in the interval of convergence of the seriesPROPERTY 2b. hhe series formed by differentiating the terms of series (1) converges to f ′(x ) for each x within the radius of convergence of (1); that is,f′(x )=∑ka k (x −a)k−1=a 1+2a 2(x −a )+⋯+na n (x −a )n−1+⋯∞k=0 (2) Note that power series (1) and its derived series (2) have the same radius of convergence but not necessarily the same interval of convergenceExample 38Let f (x )=∑x kk k+1∞k=1=x 1∙2+x 22∙3+x 33∙4+⋯+x n n n+1+⋯Find the intervals of convergence of the power series for f(x) and f ′(x ) Solution:lim n→∞|x n+1(n +1)(n +2)∙n (n +1)x n |=|x | also,f (1)=11∙2+12∙3+13∙4+⋯+1n ∙(n +1)+⋯ andf (−1)=−11∙2+12∙3−⋯+(−1)n n ∙(n +1)+⋯Hence, the power series for f converges if −1≤x ≤1 For the derivative f ′(x )=∑x k−1k+1∞k=1=12+x3+x 24+⋯+x n−1n+1+⋯lim n→∞|x n n +2∙n +1x n−1|=|x | also,f ′(1)=12+13+14+⋯ andf ′(−1)=12−13+14−⋯ Hence, the power series for f ′ converges if −1≤x <1hhus, the series given for f(x) and f ′(x ) have the same radius of convergence, but their intervals of convergence differPROPERTY 2c. hhe series obtained by integrating the terms of the given series (1) convergesto ∫f (t )dt xa for each x within the interval of convergence of (1); that is,∫f (t )dt =a 0(x −a )+a 1(x −a )22+a 2(x −a )33+⋯+a n (x −a )n+1n +1+⋯xa=∑a k (x −a )k+1k +1∞k=0Example 39Let f (x )=1(1−x )2 Show that the power series for ∫f (x )dx converges for all values of x in the interval of convergence of the power series for f(x)Solution:Obtain a series for 1(1−x )2 by long divisionhhen,1(1−x)2=1+2x+3x2+⋯+(n+1)x n+⋯It can be shown that the interval of convergence is −1<x<1 hhen by Property 2c∫1(1−x)2dx=∫[1+2x+3x2+⋯+(n+1)x n+⋯]dx11−x=c+x+x2+x3+⋯+x n+1+⋯Since when x=0we see that c=1, we have11−x=1+x+x2+x3+⋯+x n+⋯Note that this is a geometric series with ratio r=x and with a=1; if |x|<1, its sum isa 1−r =11−xC3. Finding a Power Series for a Function: Taylor and Maclaurin Series函数幕级数的展开: 泰勒级数和麦克劳林级数If a function f(x)is representable by a power series of the formc0+c1(x−a)+⋯+c n(x−a)n+⋯On an interval |x−a|<r, then the coefficients are given byc n=f(n)(a)n!, hhe seriesf(x)=f(a)+f′(a)(x−a)+f"(a)2!(x−a)2+⋯+f(n)(a)n!(x−a)n+⋯is called the Taylor series of the function f about the number a hhere is never more than one power series in (x−a)for f(x) It is required that the function and all its derivatives exist at x=a if the function f(x)is to generate a Taylor series expansionWhen a=0we have the special seriesf(x)=f(0)+f′(0)x+f"(0)2!x2+⋯+f(n)(0)n!x n+⋯called the Maclaurin series of the function f; this is the expansion of f about x=0Example 40Find the Maclaurin series for f(x)=e xSolution:Here f′(x)=e x,…,f(n)(x)=e x,…, for all n hhenf′(0)=1,…,f(n)(0)=1,…for all n, making the coefficients c n=1n!:e x=1+x+x22!+x33!+⋯+x nn!+⋯Example 41Find the Maclaurin expansion for f(x)=sin x Solution:hhus,sin x=x−x33!+x55!−⋯+(−1)n−1x2n−1(2n−1)!+⋯Example 42Find the Maclaurin expansion for f(x)=11−xSolution:hhus,11−x=1+x+x2+x3+⋯+x n+⋯Note that this agrees exactly with the power series in x obtained by different methods in Example39Example 43Find the Maclaurin expansion for f(x)=ln x about x=1Solution:hhus,ln x=(x−1)−(x−1)22+(x−1)33−(x−1)44+⋯+(−1)n−1(x−1)nnFUNCTIONS THAT GENERATE NO SERIES 不能级数展开的函数Note that the following functions are among those that fail to generate a specific series in (x−a)C4. Approximating Functions with Taylor and Maclaurin Polynomials 泰勒多项式和麦克劳林多项式的近似函数hhe function f(x)at the point x=a is approximated by a Taylor polynomial P n(x)of order n:f(x)≈P n(x)=f(a)+f′(a)(x−a)+f"(a)2!(x−a)2+⋯+f(n)(a)n!(x−a)nhhe haylor polynomial P n(x)and its first n derivatives all agree at a with f and its firstn derivatives hhe order of a haylor polynomial is the order of the highest derivative, which is also the polynomial’s last termIn the special case where a =0 , the Maclaurin polynomial of order n that approximates f(x) isP n (x )=f (0)+f ′(0)x +f"(0)2!x 2+⋯+f (n)(0)n!x nhhe haylor polynomial P 1(x) at x =0 is the tangent-line approximation to f(x) near zero given byf (x )=f (0)+f ′(0)xlt is the “best” linear approximation tof at 0, discussed at length in Chapter 4 §LA N OTE ON O RDER(泰勒多项式的阶数) AND D EGREE(泰勒多项式的级数)A haylor polynomial has degree(级数) n if it has powers of (x −a ) up through the n th If f (n)(a)=0, then the degree of P n (x) is less than n Note, for instance, in Example 45, that the second-order polynomial P 2(x) for the function sin x (which is identical with P 1(x)) is x +0∙x 22!, or just x , which has degree 1, not 2Example 44Find the haylor polynomial of order 4 at 0 for f (x )=e −x Use this to approximate f(0.25)Solution:hhe first four derivatives are −e −x ,e −x ,−e −x ,and e −x ; at a =0 , these equal −1,1,−1,and 1, respectively hhe approximating haylor polynomial of order 4 is thereforee −x ≈1−x +12!x 2−13!x 3+14!x 4 With x =0.25 we havee −0.25≈1−0.25+12!(0.25)2−13!(0.25)3+14!(0.25)4≈0.7788 hhis approximation of e −0.25 is correct to four placesIn Figure 10-2 we see the graphs of f(x) and of the haylor polynomials:Figure 10- 1P0(x)=1;P1(x)=1−x;P2(x)=1−x+x2;P3(x)=1−x+x22!−x33!;P4(x)=1−x+x22!−x33!+x44!Notice how closely P4(x)hugs f(x)even as x approaches 1Since the series can be shown to converge for x>0by the Alternating Series hest, the error in P4(x)is less than themagnitude of the first omitted term, x55!, or 1120at x=1 In fact, P4(1)=0.375to three decimalplaces, close to e−1≈0.368Example 45(a) Find the haylor polynomials P1,P3,P5, and P7at x=0for f(x)=sin x(b) Graph f and all four polynomials in [−2π,2π]×[−2,2](c) Approximate sinπ3using each of the four polynomialsSolution:P1(x)=x;P3(x)=x−x3 3!;P5(x)=x−x33!+x55!;P7(x)=x−x33!+x55!−x77!(b) Figure 10-3a shows the graphs of sin x and the four polynomials In Figure 10-3b we see graphs only of sin x and P7(x), to exhibit how closely P7“follows” the sine cruve。