数学专业英语－Sequences and SeriesSeries are a natural continuation of our study of functions. In the previous cha pter we found howto approximate our elementary functions by polynomials, with a certain error te rm. Conversely, one can define arbitrary functions by giving a series for them. We shall see how in the sections below.In practice, very few tests are used to determine convergence of series. Esse ntially, the comparision test is the most frequent. Furthermore, the most import ant series are those which converge absolutely. Thus we shall put greater emp hasis on these.Convergent SeriesSuppose that we are given a sequcnce of numbersa1,a2,a3…i.e. we are given a number a n, for each integer ｎ＞1.We form the sumsS n=a1+a2+…+a nIt would be meaningless to form an infinite suma1+a2+a3+…because we do not know how to add infinitely many numbers. However, if ou r sums S n approach a limit as n becomes large, then we say that the sum of our sequence converges, and we now define its sum to be that limit.The symbols∑a=1 ∞a nwill be called a series. We shall say that the series converges if the sums app roach a limit as n becomes large. Otherwise, we say that it does not converge, or diverges. If the seriers converges, we say that the value of the series is∑a=1∞=lim a→∞S n=lim a→∞(a1+a2+…+a n)In view of the fact that the limit of a sum is the sum of the limits, and other standard properties of limits, we get:THEOREM 1. Let{ a n}and { b n}(n=1,2,…)be two sequences and assume that the series∑a=1∞a n∑a=1∞b nconverge. Then ∑a=1∞(a n + b n ) also converges, and is equal to the sum of the two series. If c is a number, then∑a=1∞c a n=c∑a=1∞a nFinally, if s n=a1+a2+…+a n and t n=b1+b2+…+b n then∑a=1∞a n ∑a=1∞b n=lim a→∞s n t nIn particular, series can be added term by term. Of course , they cannot be multiplied term by term.We also observe that a similar theorem holds for the difference of two serie s.If a series ∑a n converges, then the numbers a n must approach 0 as n beco mes large. However, there are examples of sequences {an} for which the serie s does not converge, and yet lim a→∞a n=0Series with Positive TermsThroughout this section, we shall assume that our numbers a n are ＞0. Then t he partial sumsS n=a1+a2+…+a nare increasing, i.e.s1＜s2 ＜s3＜…＜s n＜s n+1＜…If they are approach a limit at all, they cannot become arbitrarily large. Thus i n that case there is a number B such thatS n< Bfor all n. The collection of numbers {s n} has therefore a least upper bound ,i.e. there is a smallest number S such thats n<Sfor all n. In that case , the partial sums s n approach S as a limit. In other wo rds, given any positive number ε>0, we haveS –ε< s n < Sfor all n .sufficiently large. This simply expresses the fact that S is the least o f all upper bounds for our collection of numbers s n. We express this as a theo rem.THEOREM 2. Let{a n}(n=1,2,…)be a sequence of numbers>0 and letS n=a1+a2+…+a nIf the sequence of numbers {s n} is bounded, then it approaches a limit S , wh ich is its least upper bound.Theorem 3 gives us a very useful criterion to determine when a series with po sitive terms converges:THEOREM 3. Let∑a=1∞a n and∑a=1∞b n be two series , with a n>0 for all n an d b n>0 for all n. Assume that there is a number c such thata n< cb nfor all n, and that∑a=1∞b n converges. Then ∑a=1∞a n converges, and∑a=1∞a n ≤c∑a=1∞b nPROOF. We havea1+…+a n≤cb1+…+cb n=c(b1+…+b n)≤c∑a=1∞b nThis means that c∑a=1∞b n is a bound for the partial sums a1+…+a n.The least u pper bound of these sums is therefore ≤c∑a=1∞b n, thereby proving our theore m.Differentiation and Intergration of Power Series.If we have a polynomiala0+a1x+…+a n x nwith numbers a0,a1,…,a n as coefficients, then we know how to find its derivati ve. It is a1+2a2x+…+na n x n–1. We would like to say that the derivative of a ser ies can be taken in the same way, and that the derivative converges whenever the series does.THEOREM 4. Let r be a number >0 and let ∑a n x n be a series which conv erges absolutely for ∣x∣<r. Then the series ∑na n x n-1also converges absolutel y for∣x∣<r.A similar result holds for integration, but trivially. Indeed, if we have a series ∑a=1∞a n x n which converges absolutely for ∣x∣<r, then the series∑a=1∞a n/n+1 x n+1=x∑a=1∞a n x n∕n+1has terms whose absolute value is smaller than in the original series.The preceding result can be expressed by saying that an absolutely converge nt series can be integrated and differentiated term by term and and still yields an absolutely convergent power series.It is natural to expect that iff (x)=∑a=1∞a n x n,then f is differentiable and its derivative is given by differentiating the series t erm by term. The next theorem proves this.THEOREM 5. Letf (x)=∑a=1∞a n x nbe a power series, which converges absolutely for∣x∣<r. Then f is differentia ble for ∣x∣<r, andf′(x)=∑a=1∞na n x n-1.THEOREM 6. Let f (x)=∑a=1∞a n x n be a power series, which converges abso lutely for ∣x∣<r. Then the relation∫f (x)d x=∑a=1∞a n x n+1∕n+1is valid in the interval ∣x∣<r.We omit the proofs of theorems 4,5 and 6.Vocabularysequence 序列positive term 正项series 级数alternate term 交错项approximate 逼近,近似 partial sum 部分和elementary functions 初等函数 criterion 判别准则(单数)section 章节 criteria 判别准则(多数)convergence 收敛(名词) power series 幂级数convergent 收敛(形容词) coefficient 系数absolute convergence 绝对收敛 Cauchy sequence 哥西序列diverge 发散radius of convergence 收敛半径term by term 逐项M-test M—判别法Notes1. series一词的单数和复数形式都是同一个字.例如:One can define arbitrary functions by giving a series for them(单数)The most important series are those which converge absolutely(复数)2. In view of the fact that the limit of a sum of the limits, and other standard properties of limits, we get:Theorem 1…这是叙述定理的一种方式: 即先将事实说明在前面,再引出定理. 此句用in view of the fact that 说明事实,再用we get 引出定理.3. We express this as a theorem.这是当需要证明的事实已再前面作了说明或加以证明后,欲吧已证明的事实总结成定理时,常用倒的一个句子,类似的句子还有(参看附录Ⅲ):We summarize this as the following theorem; Thus we come to the following theorem等等.4. The least upper bound of these sums is therefore ≤c∑a=1∞b n, thereby proving our theorem.最一般的定理证明格式是”给出定理…定理证明…定理证毕”,即thereby proving our theorem;或we have thus proves the theorem或This completes the proof等等作结尾(参看附录Ⅲ).5. 本课文使用较多插入语.数学上常见的插入语有:conversely; in practice; essentially; in particular; ind eed; in other words; in short; generally speaking 等等.插入语通常与句中其它成份没有语法上的关系,一般用逗号与句子隔开,用来表示说话者对句子所表达的意思的态度.插入语可以是一个词,一个短语或者一个句子.ExerciseⅠ. Translate the following exercises into Chinese:1. In exercise 1 through 4,a sequence f (n) is defined by the formula given. In each case, (ⅰ)Determine whether the sequence (the formulae are omitted).2. Assume f is a non–negative function defined for all x>1. Use the methodsuggested by the proof of the integral test to show that∑k=1n-1f(k)≤∫1n f(x)d x ≤∑k=2n f(k)Take f(x)=log x and deduce the inequalitiesc•n n•c-n< n！<c•n n+1•c-nⅡ. The proof of theorem 4 is given in English as follows(Read the proof through and try to learn how a theorem is proved, then translate this proof into Chinese ):Proof of theorem 4 Since we are interested in the absolute convergence. We may assume that a n>0 for all n. Let 0<x<r, and let c be a number such that x<c<r. Recall that lim a→∞n1/n=1.We may write n a n x n =a n(n1/n x)n. Then for all n sufficiently large, we conclude that n1/n x<c. This is because n1/n comes arbitrarily close to x and x<c. Hence for all n sufficiently large, we have na n x n<a n c n. We can then compare the series ∑nax n with∑a n c n to conclude that∑na n x n converges. Since∑na n x n-1=1n/x∑na n x n, we have proved theorem 4.Ⅲ. Recall from what you have learned in Calculus about (ⅰ) Cauchy sequence and (ⅱ) the radius of c onvergence of a power series.Now give the definitions of these two terms respectively.Ⅳ. Translate the following sentences into Chinese:1. 一旦我们能证明,幂级数∑a n z n在点z=z1收敛,则容易证明,对每一z1∣z∣<∣z1∣,级数绝对收敛;2. 因为∑a n z n在z=z1收敛,于是,由weierstrass的M—判别法可立即得到∑a n z n在点z,∣z∣<z1的绝对收敛性;3. 我们知道有限项和中各项可以重新安排而不影响和的值,但对于无穷级数,上述结论却不总是真的。
