北京邮电大学国际学院高等数学(下)幻灯片讲义(无穷级数)-Lecture 1.pdf

1 Lecture 1 1 Series of Constant Terms A Bouncing Ball Drop a ball from H meters above a flat surface Each time the ball hits the surface after falling a distance h it rebounds a distance rh where r is positive but less than 1 find the total distance the ball travels up and down It is easy to see that the total distance is 23 222sHHrHrHr The question is that how to calculate the 2 q sum with infinite terms It is well known that if there has finite terms 23 2222 2 1 1 n n n sHHrHrHrHr Hrr H r It is clear that 2 1lim 2 lim 11 n n n n Hrr Hr sHH rr Repeating Decimals Express the repeating decimal 5 23 23 23 as the ratio of two integers Solution 23 232323 5 232323 5 100 100 100 2 2311 3 2311 51 100100100 23123518 55 100 0 999999 Infinite Series If the terms of the sequence are numbers then the series is called a The following representation 12n aaa or 1 n n a which consists of all the terms of a sequence an and connected successively by the plus sign is called an infinite series or simply a series and anis called the general term of the series Definition 4 q series of constant terms If there is little chance of confusion it may be called simply a series It is convenient to use sigma notation to write the series as 1 n n a 1 k k a or n a A useful shorthand when summation from 1 to is understood Infinite Series A finite sum of real numbers always produces a real number but an infinite sum of real numbers is something else entirely This is why we need a careful definition of infinite series We begin by asking how to assign meaning to an expression like 1111 1 5 1 24816 The way to do so is not to try to add all the terms at once we cannot but rather to add the terms one at a time from the beginning and look for a pattern in how these partial sum grow Infinite Series First 1 1s 21 Second 2 1 1 2 s 1 2 2 Partial SumValue 1 1111 1 24816 n n a 6 2 2 Third 3 11 1 24 s 1 2 4 nth 1 111 1 242 nn s 1 1 2 2n 2 Partial Sum Definition The sum of the first nterms of the series 12 1 1 2 n nnk k Saaaan is called the partial sum of the series 7 The partial sum of the series form a sequence 11212 1 n nk k sasaasa of real numbers each defined as a finite sum Convergence and Divergence Definition Convergence and Divergence of a series If the sequence of partial sums of a series 1 k k a has a limit Sas napproaching infinite we say that the series converges to the sum S and we write 123 1 nk k aaaaaS 8 1k Otherwise we say that the series diverges The deference between the sum and the partial sum of the series 1 nnk k n RSSa is called the remainder of the series Geometric Series A series of the form 21 0 0 nn n aaqaqaqaqa is called a geometric series or series of equal ratios It is easy to see that the partial sum of the series is 21 1 1 1 n n n aq q Saaqaqaqq 9 1 naq Hence when 1 q 1 and 121 1 nkn k aaaaa 16 1nn k Conversely if n n ka converges for any k 1 then 1 n n a converges Theorem Adding or Deleting Terms Adding deleting or changing any finite number of terms of a series does not change the convergence or divergence of the series Properties of Series Whenever we have two convergent series we can add them term by term subtract them term by term or multiply them by constants to make a new convergent series Theorem Properties of Combining Series If n aA and n bB are convergent series then 1 SR l bbAB 17 1 Sum Rule nnnn ababAB 2 Difference Rule nnnn ababAB 3 Constant Multiple Rule nn kakakA any number k 4 Order Rule If nn abnN then 11 nn nn ab Properties of Series Theorem Associative Property If a series converges then its sum is not changed when we add arbitrarily some brackets among the terms of the series so long as we do not change the order of the terms 18 12345 aaaaa Then limlim mn mn ss 21 s 52 s 93 s nm s 4 Cauchy s convergence principle Theorem Cauchy s convergence principle The necessary and sufficient condition for a series to be convergent is that0 NN such that Np inequality 12 nnnp aaa N This theorem is a very powerful tool which is used to justify the convergence of a series Cauchy s convergence principle ExampleProve that the series 2 1 1 n n converges Proof N n p we have 222 111 1 2 nnnp 111 20 111 1 1 2 1 n nnnnpnp 111111 1121nnnnnpnp 111 nnpn Hence for any given positive take 1 N and then the inequality Cauchy s convergence principle ExampleProve that the series 2 1 1 n n converges Proof continued 111 21 222 1 2 nnnp Then by the Cauchy s convergence principle the series converges Series of Non negative Terms Given a series n a we have two questions 1 Does the series converge 2 If it converges what is the sum From now on we study series that do not have negative terms This kind of series is also called as series of positive terms The reason for this restriction is that the partial sums of these series 22 The reason for this restriction is that the partial sums of these series form non decreasing sequences and non decreasing sequences that are bounded from above always converge Theorem Necessary and sufficient condition A series of nonnegative terms is convergent iff its partial sum is bounded above Integral Test We introduce the Integral Test with an example ExampleShow that the series 22 1 11111 1 4916 n nn converges SolutionWe determine the convergence of this series by comparing 23 it with 2 1 1 dx x To carry out the comparison we think of the terms of the series as values of the function 2 1 f x x and interpret these value as area of rectangles under the curve 2 1 y x Integral Test continued 2222 1111 123 n s n 1 2 3 ffff n 2 1 1 1 n fdx x 2 1 1 1dx x It is easy to see that 24 112 1 Does converge 1 1 nn n Example 26 We have 3 2 11 1 lim k k dxxdx xx 1 2 1 lim2 k k x 2 lim22 k k Since the integral converges so must the series p Series and Harmonic Series The Integral Test can be used to settle the question of convergence for any series of the form p 0 a real constant Such series are called p series 1 1 p n n The p series 11111 27 1 123 ppppp n nn pa real constant converges if p 1 and diverges if p 1 The p series with p 1 is the harmonic series and it is probably the most famous divergent series in mathematics The Comparison Test The Direct Comparison Test Let n a and be two series of positive terms and for all b Ifdiverges thendivergesa a If converges then converges n a n b nN n b nn ab for some integer N b 28 To apply the Direct Comparison Test to a series we need not include the early terms of the series We can start the test with any index N provided that we include all the terms of the series being tested from there on b If diverges then diverges n a n b Applying the Direct Comparison Test Does the following series converge 211111 51 3723 4 k SolutionWe ignore the first four terms and compare the remaining terms with those of the convergent geometric series 1 n We see that 29 terms with those of the convergent geometric series 12 n n We see that 111111 23 4 248 Therefore the original series converges by the Direct Comparison Test Some Important Convergent and Divergent Series Convergent SeriesDivergent Series Geometric Series with 1 r The harmonic series 1 1 n n Geometric Series with 1 r Any series for which the n a limn n a does not exist or lim0 nn a 6 The Comparison Test The Limit Comparison Test 1 If then and both converge or bothlim 0 n n n a b nN n b a 31 3 If and diverges then diverges n a lim n n n a b n b 2 If and converges then converges n a lim0 n n n a b n b Proof of Part 1Since there exists an integer Nsuch that for0 2 2 n n a nN b 22 n n a b 3 n a 32 22 n n b 3 22 nnn bab By the Direct Comparison Test we have the conclusion Using the Limit Comparison Test a 22 11 35792121 491625 1 21 nn nn nnn Solution 22n 1 Let 2 21 21 n n a nn For nlarge we expect anto behave 1 Determine whether the series converge or diverge 33 like so we let 2 22n nn 1 n b n Since 11 1 n nn b n diverges and 2 2 2 limlim2 21 n nn n ann bnn n a diverges by part 1 of the Limit Comparison Test Using the Limit Comparison Test Solution Let 1 21 nn a likeso we let 1 1 b For nlarge we expect anto behave Since 1 b converges and b 1 11111 1371521 n n 34 like so we let 2n 2 nn b Since 112 nn nn b converges and 21 limlimlim1 211 1 2 n n nn nnn n a b n a converges by part 1 of the Limit Comparison Test Using the Limit Comparison Test Solution Let 2 1ln 5 n nn a n For nlarge we expect anto behave c 2 2 12ln213ln314ln41ln 914215 n nn n likewhich is greater thanforso we take lnlnnnn 1 b 1 3n 35 Since 11 1 n nn b n diverges and 2 2 ln limlim 5 n nn n annn bn n a diverges by part 3 of the Limit Comparison Test like which is greater than for so we take 2 nn n b n n 3n Ratio and Root Tests The Ratio Test d Alembert s Test Let be a series with positive terms and suppose that Then a the series converges if b the series diverges if c the test is inconclusive if n a 1 lim n n n a a 1 1 36 Proof Part a 1 1n n a a Let be a number between and 1 r 7 Ratio and Root Tests 2 21NNN arar a 1 m NmNmN arar a Then 1NN ara Proof of Ratio Test continued Part a 1 37 1NmNmN These inequalities show that the terms of our series after Nth term approach zero more rapidly than the terms in a geometric series with ratio 1r Ratio and Root Tests Proof of Ratio Test More precisely consider the series where for and for all n and n c nn ca 1 2 nN 2 12 m NNNNNmN cracr acr a 2 caaaarar a continued Part a 1 38 121 1 nNNNN n caaaarar a 2 121 1 NN aaaarr The geometric series converges because so converges Since also converges 2 1rr 1r and 12 MMM aaa The terms of the series do not approach zero as nbecomes infinite and the series diverges by the nth Term Test Part b 1 The two series and show that some other 1 2 1 39 e t o se esa ds ot at soe ot e 1n n 2 1n n test for convergence must be used when 1 For 2 1 1 n n 2 2 1 2 1 1 1 11 n n ann ann 1 1 n n For 1 1 1 1 11 n n ann ann Ratio and Root Tests The nth Root Test Cauchy s Test Let be a series with for and suppose that Then a the series converges if b the series diverges if c the test is inconclusive if n a limn n n a 1 1 0 n a nN 40 ExerciseWhich of the following series converge and which diverge a b 2 12 n n n 2 1 2n n n Alternating Series Definition Alternating Series A series in which the terms are alternately positive and negative is an alternating series 1 1111 1 1 n 41 2345n 1 123456 1 nn The Alternating Series Test Theorem Leibniz s Theorem The series converges if all of the following conditions are satisfied 1 the un s are all positive 2 for all for some integer N 1nn uu nN 1 1234 1 1 n n n uuuuu 42 3 0 n u ProofIf nis an even integer say then the sum of the first 2nm nterm is 21234212 mmm suuuuuu 0 0 0 222mm ss 1234522212 mmm uuuuuuuu 21m su 8 The Alternating Series Test continued If nis an odd integer say then the sum of the first nterms is Since and as 21nm 21221mmm ssu 0 n u m Since is nondecreasing and bounded from above it has a limit 2 m s 2 lim m m sL 1 43 21221mmm n 21221 0 mmm ssuLL 2 It is easy to see that the series is convergent 1 1 1 n n n lim n n sL Combining the results of equation 1 and 2 gives Absolute Convergence Definition Conditional Convergence A series that converges but does not converge absolutely converges conditionally Definition Absolute Convergence The series converges absolutely is absolutely convergent if the corresponding series of absolute values converges n a n a 44 but does not converge absolutely converges conditionally Absolute convergence is important for two reasons First we have good tests for convergence of series of positive terms Second if a series converges absolutely then it converges The Absolute Convergence Test Theorem The Absolute Convergence Test If converges then converges 1 n n a 1 n n a Proof For each n 45 nnn aaa so0 2 nnn aaa If converges then converges and by the Direct Comparison Test the nonnegative series converges The equality lets us express the series as the difference of two convergent series 1 n n a 1 2 n n a 1 nn n aa nnnn aaaa The Absolute Convergence Test Theorem The Absolute Convergence Test If converges then converges 1 n n a 1 n n a Proof continued 46 1111 nnnnnnn nnnn aaaaaaa Therefore converges 1 n n a ExampleThe