2 +1 2(1) (2) (3) = ( +1)! 2! 1 3! 1 4! 1 5! 1 1 6! = 2 = 6 2 =3.

Transcription

1 INFINITE SEQUENCES AND SERIES Sequences (a) A sequence is an ordered list of numbers It can also be defined as a function whose domain is the set of positive integers (b) The terms approach 8 as becomes large In fact, we can make as close to 8 as we like by taking sufficiently large (c) The terms become large as becomes large In fact, we can make as large as we like by taking sufficiently large (a) From Definition, a convergent sequence is a sequence for which exists Examples: {}, { } (b) A divergent sequence is a sequence for which does not exist Examples: {}, {sin } =, so the sequence is + () + () + () + 4 (4) + 5 (5) + = =, so the sequence is = = ( ), so the sequence is = =cos, so the sequence is cos cos cos cos cos 5 = {0 0 0} 7 =, so the sequence is ( +)!!! 4! 5! 6! = = ( )!+,so = ( )! + =, and the sequence is = =, + =5 Each term is defined in terms of the preceding term =5 =5() = =5 =5() =7 4 =5 =5(7) = 5 =5 4 = 5() =57 The sequence is { 7 57} 0 =6, + = = = 6 =6 = = 6 = 4 = = = 5 = 4 4 = 4 The sequence is =, + = + = + = + = = + = + = 5 4 = + = 5 +5 = 7 5 = 4 = = 9 The sequence is c 06 Cengage Learning All Rights Reserved May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part 957

2 958 CHAPTER INFINITE SEQUENCES AND SERIES =, =, + = Eachtermisdefinedintermofthetwoprecedingterms = = = 4 = = = 5 = 4 = ( ) = 6 = 5 4 = ( ) = The sequence is { }, 4, 6, 8, 0, The denominator is two times the number of the term,,so = 4 5 4,, 4, 6, 64, Thefirsttermis4 andeachtermis 4 times the preceding one, so = Thefirsttermis and each term is times the preceding one, so = 6 { } Each term is larger than the preceding term by,so = + ( ) = 5 + ( ) = we get =( ) + + Thenumeratorofthenth term is and its denominator is + Including the alternating signs, 8 { } Two possibilities are =sin and ( ) =cos 9 0 = =+ ( ) It appears that =05 +6 () ( + 6) +6 = 6 = It appears that = + ( ) ( ) + =+0=since =0 ( ) and by Theorem 6, =0 c 06 Cengage Learning All Rights Reserved May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part

3 SECTION SEQUENCES 959 = = It appears that = + + =0by (9) =+0=since It appears that the sequence does not have a it 0 0 9, which diverges by (9) since = +5 + = ( + 5 ) ( + ) = 5+ +,so 5+0 =5as Converges +0 4 = +5 + = ( + 5 ) +5 = ( + ) +,so as since +5 = and + =0+= Diverges 4 5 = = 4 ( ) =,so as since / = and = 0= Diverges 6 =+(086) +0=as since (086) =0by (9) with =086 Converges 7 = 7 = 7 =,so 7 =0by (9) with = 7 Converges c 06 Cengage Learning All Rights Reserved May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part

4 960 CHAPTER INFINITE SEQUENCES AND SERIES 8 = = + ( +) = + =as Converges +0 9 Because the natural exponential function is continuous at 0, Theorem 7 enables us to write = ( ) = 0 = Converges 0 = 4 +9 = 4 9 ( + 9 )9 = (49) (9) =0by (9) Converges 9 =0as since 4 =0 and 9 +4 = + = ( + 4 ) ( + ) = ( )+4 ( )+ 4=as since ( )=0 Converges =cos =cos =cos,so cos = as since =0 + ( +) + Converges = +4 = +4 =,so as since = and = Diverges 4 If =,then + () ( +) continuous at, by Theorem 7, (+) = = 5 ( ) = 6 + ( + ) + = = Since the natural exponential function is Converges = (0) = 0,so =0by (6) Converges + = +0 = Thus, = ( )+ + has odd-numbered terms that approach and even-numbered terms that approach as, and hence, the sequence { } is divergent ( )! 7 = ( +)! = ( )! ( +)()( )! = 0 as Converges ( +)() 8 = ln ln = ln ln + ln = ln + =as Converges 0+ ln 9 =sin This sequence diverges since the terms don t approach any particular real number as Thetermstakeon values between and Diverges 40 = tan tan tan = by (), so =0 Converges 4 = = Since = H = H =0, it follows from Theorem that =0 Converges c 06 Cengage Learning All Rights Reserved May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part

5 SECTION SEQUENCES =ln( +) ln =ln =ln + ln () = 0 as because ln is continuous Converges 4 0 cos [since 0 cos ], so since =0, cos converges to 0 by the Squeeze Theorem 44 = + =( + ) =( ) = =8,so = 8 =8 () =8 0 =8by Theorem 7, since the function () = is continuous at 0 Converges 45 = sin() = sin() sin() sin Since [where =] =, it follows from Theorem 0 + that { } converges to 46 = cos cos 0 =,so =0by (9), and =0by (6) 47 = + ln = ln +,so ln( + ) H + ln + = + ln =,sobytheorem, + = Converges 48 = ln = ln H ln, so ln =0 ln = 0 =,sobytheorem, = Converges 49 =ln( +) ln( + + +)=ln =ln ln as + + Converges Converges (ln ) H (ln )() ln H (ln ) 50 = =0,sobyTheorem, =0 Converges 5 =arctan(ln) Let() = arctan(ln ) Then () = since ln as and arctan is continuous Thus, () = Converges 5 = + += +4 += +4 + = ( +4 +) = = so 4 0 = = 4 = Converges ( 4 ) = , c 06 Cengage Learning All Rights Reserved May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part

6 96 CHAPTER INFINITE SEQUENCES AND SERIES 5 { } diverges since the sequence takes on only two values, 0 and, and never stays arbitrarily close to either one (or any other value) for sufficiently large = and = for all positive integers + =0since =0and + =0Forsufficiently large, can be made as close to 0 as we like Converges 55 =! = ( ) 56 0 =! = ( ) Theorem and Theorem 6, {( )!} converges to 0 [for ] = as,so{} diverges 4 7 [for ] = 0 as, so by the Squeeze 57 From the graph, it appears that the sequence { } = ( ) is + divergent, since it oscillates between and (approximately) To prove this, suppose that { } converges to If =,then{} converges to, + and = = But =( ),so does not exist This contradiction shows that { } diverges 58 From the graph, it appears that the sequence converges to 0 = sin sin =,so =0 By (6), it follows that =0 59 From the graph, it appears that the sequence converges to a number between 07 and 08 =arctan =arctan =arctan +4 ( +4) +4 arctan = 4 [ 0785]as 60 From the graph, it appears that the sequence converges to 5 5 = = 5 = 5 5 as Hence, 5 by the Squeeze Theorem = 0 = [continued] c 06 Cengage Learning All Rights Reserved May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part

7 SECTION SEQUENCES 96 Alternate solution: Let =( +5 ) Then ln ( +5 ) ln H ln + 5 ln 5 +5 so = ln 5 =5,andso +5 converges to 5 ln + ln 5 5 =ln5, From the graph, it appears that the sequence { } = 6 cos is + divergent, since it oscillates between and (approximately) To prove this, suppose that { } converges to If = { } converges to,and +,then = = But =cos,so does not exist This contradiction shows that { } diverges 5 ( ) From the graphs, it seems that the sequence diverges = We first prove by induction that! for all Thisisclearlytruefor =,solet () be the statement that the above is true for Wemust + show it is then true for + + = (induction hypothesis) But + + [since ( +) ( +) ], andsowegetthat + = which is ( +) Thus, we have proved our first assertion, so since diverges [by (9)], so does the given sequence { } 6 From the graph, it appears that the sequence approaches 0 0 = 5 ( ) () = () () () = 5 0 as 5 ( ) So by the Squeeze Theorem, () converges to 0 c 06 Cengage Learning All Rights Reserved May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part

8 964 CHAPTER INFINITE SEQUENCES AND SERIES 64 (a) =, + =4 for =, =4 =4 =, =4 =4 =, 4 =4 =4 =, 5 =4 4 =4 = Since the terms of the sequence alternate between and, the sequence is divergent (b) =, =4 =4 =, =4 =4 = Since all of the terms are, =and hence, the sequence is convergent 65 (a) = 000(06) =060, =60, =90, 4 = 648,and 5 =8 (b) = 000 (06), so the sequence diverges by (9) with = (a) Substitute to 6 for in =00 to get =$0, =$05, =$075, 4 =$50, =$5,and 6 =$76 (b) For two years, use = 4 for to get $ (a) We are given that the initial population is 5000,so 0 = 5000 The number of catfish increases by 8% per month and is decreased by 00 per month, so = 0 +8% 0 00 = , =08 00, and so on Thus, =08 00 (b) Using the recursive formula with 0 = 5000,weget =500, =508, =55(rounding any portion of a 68 + = catfish), 4 =545, 5 =5587,and 6 = 574, which is the number of catfish in the pond after six months + if is an even number if is an odd number When =,thefirst40 terms are, 4, 7, 5, 6,, 40, 0, 0, 5, 6, 8, 4,,, 4,,, 4,,, 4,,, 4,,, 4,,, 4,,, 4,,, 4,,, 4 When =5,thefirst40 terms are 5, 76, 8, 9, 58, 9, 88, 44,,, 4, 7, 5, 6,, 40, 0, 0, 5, 6, 8, 4,,, 4,,, 4,,, 4,,, 4,,, 4,,, 4 The famous Collatz conjecture is that this sequence always reaches, regardless of the starting point 69 If, then{ } diverges by (9), so { } diverges also, since = If then whenever = H =0,so ( ln ) ln =0, and hence { } converges 70 (a) Let = By Definition, this means that for every 0there is an integer such that whenever Thus, + whenever + It follows that + = and so + (b) If then + = also, so must satisfy = ( + ) + =0 = + 5 (since has to be nonnegative if it exists) c 06 Cengage Learning All Rights Reserved May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part

9 SECTION SEQUENCES Since { } is a decreasing sequence, + for all Because all of its terms lie between 5 and 8, { } is a bounded sequence By the Monotonic Sequence Theorem, { } is convergent; that is, { } has a it must be less than 8 since { } is decreasing, so Since { } = {cos } { }, the sequence is not monotonic The sequence is bounded since cos for all 7 = + is decreasing since + = ( +)+ = +5 + = for each The sequence is bounded since 0 for all Notethat 5 = ( +) + +( +) ,which is true for all,so{ } is decreasing Since =0and + =, the sequence is bounded + ( 0) 75 The terms of = ( ) alternate in sign, so the sequence is not monotonic The first five terms are,,, 4,and 5 Since =, the sequence is not bounded 76 Since { } = + ( ) =, the sequence is not monotonic The sequence is bounded since 5 for all 77 = Let() = Then 0 () =0 [( )+ ]= ( ), which is positive for,so is increasing on ( ) It follows that the sequence { } = {()} is increasing The sequence is bounded below by = 6 and above by, so the sequence is bounded 78 = +Let() = +Then 0 () = =( ), which is positive for,so is increasing on ( ) It follows that the sequence { } = {()} is increasing The sequence is bounded below by =, but is not bounded above, so it is not bounded, 79 For,,, =, = 4, = 78,,so = ( ) = () () = = Alternate solution: Let (We could show the it exists by showing that { } is bounded and increasing) Then must satisfy = = ( ) = 0 6= 0since the sequence increases, so = 80 (a) Let be the statement that + and is obviously true We will assume that is true and then show that as a consequence + must also be true , which is the induction hypothesis + + c 06 Cengage Learning All Rights Reserved May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part

10 966 CHAPTER INFINITE SEQUENCES AND SERIES + 9 7, which is certainly true because we are assuming that So is true for all,andso (showing that the sequence is bounded), and hence by the Monotonic SequenceTheorem, exists (b) If,then + = also, so = + =+ =0 ( +)( ) = 0 = [since can t be negative] 8 =, + = We show by induction that { } is increasing and bounded above by Let be the proposition that + and 0 Clearly is true Assume that is true Then Now + = + = + + This proves that { } is increasing and bounded above by,so=,thatis,{ } is bounded, and hence convergent by the Monotonic Sequence Theorem If,then + = also, so must satisfy = +=0 = ± But,so = =, + = We use induction Let be the statement that 0 + Clearly is true, since =( ) = Now assume that is true Then = = + Also + 0 [since + is positive] and + by the induction + hypothesis, so + is true To find the it, we use the fact that + = +=0 = ± 5 But,sowemusthave = 5 8 (a) Let be the number of rabbit pairs in the nth month Clearly == Inthenth month, each pair that is or more months old (that is, pairs) will produce a new pair to add to the pairs already present Thus, = +,sothat{ } = { }, the Fibonacci sequence (b) = + = + = =+ =+ 5 =+ If /, then and,somust satisfy =+ =0 = + [since must be positive] 5 84 (a) If is continuous, then () ( ) + = by Exercise 70(a) (b) By repeatedly pressing the cosine key on the calculator (that is, taking cosine of the previous answer) until the displayed value stabilizes, we see that c 06 Cengage Learning All Rights Reserved May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part

11 SECTION SEQUENCES (a) From the graph, it appears that the sequence! 5 converges to 0,thatis,! =0 (b) From the first graph, it seems that the smallest possible value of corresponding to =0 is 9,since 5! 0 whenever 0,but9 5 9! 0 From the second graph, it seems that for =000, the smallest possible value for is since 5! 000 whenever 86 Let 0 and let be any positive integer larger than ln() ln If,thenln() ln ln ln [since ln 0] ln( ) ln 0, and so by Definition, =0 87 Theorem 6: If =0then =0, and since,wehavethat =0by the Squeeze Theorem 88 Theorem 7: If = and the function is continuous at,then ( )=() Proof: We must show that, given a number 0, there is an integer such that ( ) () whenever Suppose 0 Since is continuous at, there is a number 0 such that () () if Since =, there is an integer such that if Suppose Then0,so ( ) () 89 To Prove: If =0and {} is bounded, then () =0 Proof: Since { } is bounded, there is a positive number such that and hence, for all Let0be given Since =0, there is an integer such that 0 if Then 0 = = = 0 = for all Since was arbitrary, ( )=0 c 06 Cengage Learning All Rights Reserved May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part

12 968 CHAPTER INFINITE SEQUENCES AND SERIES 90 (a) + + = =( +) (b) Since 0, wehave + + ( +) ( ) + ( +) ( ) + [( +) ] + (c) With this substitution, ( +) =,andso = (d) With this substitution, we get + (e) since { } is increasing, so = (f ) Since { } is increasing and bounded above by 4, 4,andso{ } is bounded and monotonic, and hence has a it by the Monotonic Sequence Theorem 9 (a) First we show that = + = + = 0 [since ] Also = ( + ) = ( ) 0 and = = 0,so Inthesame way we can show that andsothegivenassertionistruefor = Suppose it is true for =,thatis, + + Then + + = ( ) + + = = + + 0, + + = + ( )= ( + + ) 0,and + + = = , so the assertion is true for = + Thus, it is true for all by mathematical induction (b) From part (a) we have + +, which shows that both sequences, { } and { }, are monotonic and bounded So they are both convergent by the Monotonic Sequence Theorem + (c) Let = and = Then + = + = = + 9 (a) Let 0 Since =, there exists such that for Since + =, there exists such that + for Let =max{ +} and let If is even, then = where,so = If is odd, then = +,where,so = + Therefore = (b) =, =+ + = =5, =+ 5 = 7 5 =4, 4 =+ 5 = 7 =46, 5 =+ = 4 479, 6 =+ = , 7 =+ = 9 440, c 06 Cengage Learning All Rights Reserved May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part

13 SECTION SEQUENCES =+ = Notice that 5 7 and It appears that the odd terms are increasing and the even terms are decreasing Let s prove that and + by mathematical induction Suppose that Then We have thus shown, by induction, that the odd terms are increasing and the even terms are decreasing Also all terms lie between and, so both { } and { } are bounded monotonic sequences and are therefore convergent by the Monotonic Sequence Theorem Let = Then + = also We have + =+ ++( + ) =+ ( + )( + ) = 4+ + so + = 4+ Taking its of both sides, we get = =4+ = = = [since 0] Thus, = Similarly we find that + = So, by part (a), 9 (a) Suppose { } converges to Then + = + + = = = ( + ) =0 =0or = (b) + = = + + since + (c) By part (b), 0, 0, 0, etc In general, 0, so 0 =0since By (7) =0if Here = (0 ) (d) Let We first show, by induction, that if 0,then and + For =0,wehave 0 = 0 0 = 0( 0 ) 0 since 0 So Now we suppose the assertion is true for =, thatis, and + Then + = = ( )+ = ( ) 0 because So And + + = + + = +( + ) 0 since + Therefore, Thus, the assertion is true for = + It is therefore true for all by mathematical induction A similar proof by induction shows that if 0,then and { } is decreasing In either case the sequence { } is bounded and monotonic, so it is convergent by the Monotonic Sequence Theorem It then follows from part (a) that = c 06 Cengage Learning All Rights Reserved May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part

14 970 CHAPTER INFINITE SEQUENCES AND SERIES LABORATORY PROJECT Logistic Sequences To write such a program in Maple it is best to calculate all the points first and then graph them One possible sequence of commands [taking 0 = and =5 for the difference equation] is t:= t ;p(0):=/;k:=5; for j from to 0 do p(j):=k*p(j-)*(-p(j-)) od; plot([seq([t,p(t)] t=00)],t=00,p=005,style=point); In Mathematica, we can use the following program: p[0]=/ k=5 p[j_]:=k*p[j-]*(-p[j-]) P=Table[p[t],{t,0}] ListPlot[P] With 0 = and =5: With 0 = and =5: Both of these sequences seem to converge the first to about, the second to about 060 c 06 Cengage Learning All Rights Reserved May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part

15 LABORATORY PROJECT LOGISTIC SEQUENCES 97 With 0 = 7 8 and =5: With 0 = 7 8 and =5: The it of the sequence seems to depend on, but not on 0 With 0 = 7 and =: It seems that eventually the terms fluctuate between two values (about 05 and 08 in this case) c 06 Cengage Learning All Rights Reserved May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part

16 97 CHAPTER INFINITE SEQUENCES AND SERIES With 0 = 7 8 and =4: With 0 = 7 8 and =45: From the graphs above, it seems that for between 4 and 5, the terms eventually fluctuate between four values In the graph below, the pattern followed by the terms is Note that even for =4 (as in the first graph), there are four distinct branches ; even after 000 terms, the first and third terms in the pattern differ by about 0 9, while the first and fifth terms differ by only 0 0 With 0 = 7 and =48: 8 c 06 Cengage Learning All Rights Reserved May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part

17 SECTION SERIES =05, =7 0 =050, =7 0 =075, =9 0 =0749, =9 0 =05, =999 From the graphs, it seems that if 0 is changed by 000, the whole graph changes completely (Note, however, that this might be partially due to accumulated round-off error in the CAS These graphs were generated by Maple with 00-digit accuracy, and different degrees of accuracy give different graphs) There seem to be some some fleeting patterns in these graphs, but on the whole they are certainly very chaotic As increases, the graph spreads out vertically, with more extreme values close to 0 or Series (a) A sequence is an ordered list of numbers whereas a series is the sum of a list of numbers (b) A series is convergent if the sequence of partial sums is a convergent sequence A series is divergent if it is not convergent =5means that by adding sufficiently many terms of the series we can get as close as we like to the number 5 = In other words, it means that =5,where is the th partial sum, that is, = c 06 Cengage Learning All Rights Reserved May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part

18 974 CHAPTER INFINITE SEQUENCES AND SERIES 4 = = [ (08) ] (08) = (0) = 4 + ( ) (4 +) 4+ = = 4 5 For = 4 +, = 4 + = = 4 + = =05, = + = =055, = + 056, 4 = , 5 = , 6 = , 7 = , and 8 = It appears that the series is convergent 6 For =, = = = =, = + =+ 797, = + 487, 4 = , 5 = , 6 = , 7 = , and 8 = It appears that the series is divergent 7 For = sin, =sin = =sin 0845, = , = + 899, 4 = + 4 5, 5 = , 6 = , 7 = , and 8 = It appears that the series is divergent 8 For 9 = ( ), =( )!! = =! =, = + =! =05, = + =05+! 06667, 4 = + 4 =065, 5 = , 6 = , 7 = , and 8 = It appears that the series is convergent From the graph and the table, it seems that the series converges to In fact, it is a geometric series with = 4 and = 5,soitssumis = Note that the dot corresponding to =is part of both { } and { } ( 5) = 4 = 4 = 5 TI-86 Note: To graph { } and { }, set your calculator to Param mode and DrawDot mode (DrawDot is under GRAPH, MORE, FORMT (F)) Now under E(t)= make the assignments: xt=t, yt=/(-5)ˆt, xt=t, yt=sum seq(yt,t,,t,) (sum and seq are under LIST, OPS (F5), MORE) Under WIND use,0,,0,0,,-,, to obtain a graph similar to the one above Then use TRACE (F4) to see the values c 06 Cengage Learning All Rights Reserved May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part

19 SECTION SERIES The series cos diverges, since its terms do not approach 0 = The series = diverges, since its terms do not approach From the graph and the table, we see that the terms are getting smaller and may approach 0, and that the series approaches a value near 6 The series is geometric with =49 and =07,soitssumis = = = 49 0 =6 c 06 Cengage Learning All Rights Reserved May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part

20 976 CHAPTER INFINITE SEQUENCES AND SERIES From the graph and the table, we see that the terms are getting smaller and may approach 0, and that the series may approach a number near Using partial fractions, we have = = = = = As,,so = = From the graph and the table, we see that the terms are getting smaller and may approach 0,and that the series may approach a number near sin = sin = sin sin + sin + sin + + sin +sin + sin sin + =sin sin + As, sin sin sin sin 0 = sin,so + sin sin =sin = c 06 Cengage Learning All Rights Reserved May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part

21 SECTION SERIES (a) + =,sothesequence {} is convergent by () (b) Since = 6=0,theseries is divergent by the Test for Divergence = 6 (a) Both and represent the sum of the first terms of the sequence { },thatis,theth partial sum = = (b) = =, which, in general, is not the same as = = = terms is a geometric series with ratio = 4 Since = 4 9, the series diverges is a geometric series with ratio Since =, the series converges to = 4 4 = is a geometric series with ratio = Since = 0 5 5, the series converges to = 0 ( 5) = 0 65 = 50 6 = is a geometric series with ratio = 05 = Since = 4 4, the series converges 4 to = 4 = 4 = 8 = (07) is a geometric series with first term =and ratio =07 Since =07, the series converges to = 07 = 07 = (00) = =5 The latter series is geometric with = and ratio = Since =, itconvergesto = = = Thus, the given series converges to 5 = 5 = ( ) 4 = 4 The latter series is geometric with =and ratio = 4 Since =, it 4 4 = converges to ( 4) = 4 Thus, the given series converges to 7 =0 + ( ) = = = is a geometric series with ratio = Since =, the series diverges 7 5 = diverges 6 = = ( ) 6 6 =6 = 6 is a geometric series with ratio = 6 Since = [ ],theseries 6 c 06 Cengage Learning All Rights Reserved May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part

22 978 CHAPTER INFINITE SEQUENCES AND SERIES 6 = 6 = = 6( ) = = 4 is a geometric series with ratio = 4 Since = 4,theseries 7 diverges = = = = This is a constant multiple of the divergent harmonic series, so it diverges = , which are both convergent geometric series with sums 9 = 8 and 9 9 = 4, so the original series converges and its sum is + = diverges by the Test for Divergence since + = 6=0 = = = diverges by the Test for Divergence since +5 = = +5 =6= = = The latter series is geometric with = and ratio = 4 Since = 4, 4 it converges to = Thus, the given series converges to () = 9 4 [( 0) +(06) ]= ( 0) + (06) 4 5 = = = = = 0 = ( 0) + 06 = = 7 diverges by the Test for Divergence since since 4 [sum of two geometric series] 4+ = 4+0 = 4 6=0 +4 diverges by the Test for Divergence since = (sin 00) is a geometric series with first term = sin 00 [ 0506]andratio =sin00since,theseries = 6 converges to = + sin 00 sin diverges by the Test for Divergence since + = =6= 0 +0 c 06 Cengage Learning All Rights Reserved May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part

23 7 = + ln diverges by the Test for Divergence since + ln + + =ln =ln + + 6=0 SECTION SERIES ( ) = =0 =0 is a geometric series with first term = 0 =and ratio = series converges to = 44 arctan diverges by the Test for Divergence since arctan = 6=0 = = = = Since,the diverges because = diverges (If it converged, then would also converge by 4 4 Theorem 8(i), but we know from Example 9 that the harmonic series difference 5 + = we have just seen that = = = 5 must converge (since = = diverges, so the given series must also diverge = diverges) If the given series converges, then the 5 is a convergent geometric series) and equal =,but = is a geometric series with first term = and ratio = Since =, the series converges = to = = ByExample8, + = ( +) = = = + = = = Thus, by Theorem 8(ii), ( +) ( +) = += + = diverges by the Test for Divergence since 4 Using partial fractions, the partial sums of the series = = = = are ( )( +) = = = H H = 6= 0 + This sum is a telescoping series and =+ Thus, + = = c 06 Cengage Learning All Rights Reserved May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part

24 980 CHAPTER INFINITE SEQUENCES AND SERIES 44 For the series ln +, = =(ln ln ) + (ln ln ) + (ln ln 4) + +[ln ln( +)]=ln ln( +)= ln( +) Thus, =, so the series is divergent 45 For the series Thus, = ( +), = ( +) = = = = 46 For the series ( +) =4 + [telescoping series] [using partial fractions] The latter sum is + = [telescoping series] + =+ + = Converges 6 = =4 = = [telescoping series] Thus, + 4 = 0= + 4 Converges =4 47 For the series (+), = = (+) =( )+( )+ + (+) = (+) = [telescoping series] Thus, (+) (+) = 0 = Converges = 48 Using partial fractions, the partial sums of the series = are = = ( )( +) = = + + = + = + + = = Thus, = + Note: In three consecutive expressions in parentheses, the rd term in the first expression plus the nd term in the second expression plus the st term in the third expression sum to = = + 4 c 06 Cengage Learning All Rights Reserved May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part

25 SECTION SERIES (a) Many people would guess that, but note that consists of an infinite number of 9s (b) = = ,000 + = =0 Itssumis 09 0 = 09 =,thatis, = 09 = 9, which is a geometric series with =09 and 0 (c) The number has two decimal representations, and (d) Except for 0, all rational numbers that have a terminating decimal representation can be written in more than one way For example, 05 can be written as as well as =, =(5 ) =(5 ) =()=, =(5 ) =()=6, 4 =(5 4) =(6)=6, 5 =(5 5) 4 =0, and all succeeding terms equal 0 Thus, = = 4 = =++6+6= = is a geometric series with = 0 0 and = 0 Itconvergesto = 80 0 = = is a geometric series with = and = 00 Itconvergesto = = = Now is a geometric series with = 06 0 and = 0 Itconvergesto = = = Thus, 56 = = = = Now is a geometric series with = 05 0 and = 0 Itconverges to = 50 0 = = Thus,05 = 0+ = = 0,04 = = Now is a geometric series with = and 6 = 0 Itconvergesto = = = ,000 = 7,000 Thus, 4567 = 4 + 7,000 = ,000 = 45,658 7, ,000 = 45,679 7, = 5 + 7,58 + 7,58 + Now 7,58 + 7,58 + is a geometric series with = 7,58 and = 0 5 Itconvergesto = 7, = 7, , = 7,58 99,999 =,786, Thus, 5758 = 5 +,786, = 66,665, +,786, = 90,45, c 06 Cengage Learning All Rights Reserved May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part

26 98 CHAPTER INFINITE SEQUENCES AND SERIES ( 5) = ( 5) is a geometric series with = 5, so the series converges = = 5 5,thatis, 5 5 In that case, the sum of the series is = 5 ( 5) = 5 +5 ( +) is a geometric series with = +, so the series converges + = + In that case, the sum of the series is = + ( +) = + ( ) 59 = is a geometric series with =, so the series converges =0 =0 5 In that case, the sum of the series is = = = ( ) 5 ( 4) ( 5) = [ 4( 5)] is a geometric series with = 4( 5), so the series converges =0 =0 4( 5) In that case, the sum of the series is = [ 4( 5)] = 4 9 = is a geometric series with =, so the series converges =0 =0 or In that case, the sum of the series is = = sin = sin is a geometric series with = sin, so the series converges =0 =0 sin sin, which is true for all Thus, the sum of the series is = (sin ) = sin = ( ) is a geometric series with =, so the series converges =0 =0 0 0 In that case, the sum of the series is = 64 Because 0 and ln is continuous, we have + ln =ln=0 We now show that the series ln + = + ln = [ln( +) ln ] diverges = = = =(ln ln ) + (ln ln ) + +(ln( +) ln ) =ln( +) ln = ln( +) As, =ln( +), so the series diverges c 06 Cengage Learning All Rights Reserved May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part

27 SECTION SERIES After defining, Weuseconvert(f,parfrac); in Maple, Apart in Mathematica, or Expand Rational and Simplify in Derive to find that the general term is + + ( + ) = = = ( +) = Sothenth partial sum is ( +) = ( +) ( +) The series converges to = This can be confirmed by directly computing the sum using sum(f,n=infinity); (in Maple), Sum[f,{n,,Infinity}] (in Mathematica), or Calculus Sum (from to )andsimplify (in Derive) 66 See Exercise 65 for specific CAS commands = 4( ) + 4( +) 6( ) 6( +) + Sotheth partial sum is 4 = 4 = = The terms with denominator 5 or greater cancel, except for a few terms with in the denominator So as, 4 + = = For =, =0since =0For, = = = ( ) ( ) ( +)( ) = = + ( ) + ( +) ( +) Also, + = 68 = = = 5 For 6=, = = ( ) ( ) = + Also, =because H ln =0 = ( ) = = 69 (a) The quantity of the drug in the body after the first tablet is 00 mg After the second tablet, there is 00 mg plus 0% of the first 00-mg tablet;thatis, (00) = 0 mg After the third tablet, the quantity is (0) or, equivalently, (00) + 00(00) Either expression gives us 4 mg (b) From part (a), we see that + = (c) =00+00(00) + 00(00) + +00(00) = 00(00) = [geometric with = 00 and =00] The quantity of the antibiotic that remains in the body in the long run is = = 00 = 5 mg 45 c 06 Cengage Learning All Rights Reserved May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part

28 984 CHAPTER INFINITE SEQUENCES AND SERIES 70 (a) The concentration of the drug after the first injection is 5mgL Reduced by 90% isthesameas0% remains, so the concentration after the second injection is 5+00(5) = 65 mgl The concentration after the third injection is 5+00(65), or, equivalently, 5+5(00) + 5(00) Either expression gives us 665 mgl (b) =5+5(00) +5(00) + +5(00) = 5(00) = By (), = 5[ (00) ] 00 [geometric with =5 and =00] = 5 09 [ (00) ]= 5 [ (00) ]mgl 5 (c) The iting value of the concentration is [ (00) ]= 5 ( 0) = 5 mgl 7 (a) The quantity of the drug in the body after the first tablet is 50 mg After the second tablet, there is 50 mg plus 5% of the first 50- mg tablet, that is, [ (005)] mg After the third tablet, the quantity is [ (005) + 50(005) ]=57875 mg After tablets, the quantity (in mg) is (005) + +50(005) We can use Formula to write this as 50( 005 ) 005 = ( 005 ) (b) The number of milligrams remaining in the body in the long run is 000 ( ) = 000 ( 0) 57895, 9 only 00 mg more than the amount after tablets 7 (a) The residual concentration just before the second injection is ; before the third, + ; before the ( +)st, This sum is equal to [Formula ] (b) The iting pre-injection concentration is = ( 0) = (c), so the minimal dosage is = 7 (a) The first step in the chain occurs when the local government spends dollars The people who receive it spend a fraction of those dollars, that is, dollars Those who receive the dollars spend a fraction of it, that is, dollars Continuing in this way, we see that the total spending after transactions is = = ( ) by () (b) ( ) = ( )= since 0 =0 = [since + =] = [since =] If =08,then = =0 and the multiplier is = =5 c 06 Cengage Learning All Rights Reserved May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part

29 SECTION SERIES (a) Initially, the ball falls a distance, then rebounds a distance, falls, rebounds,falls, etc The total distance it travels is = = = + = meters (b) From Example in Section, we know that a ball falls meters in seconds, where is the gravitational acceleration Thus, a ball falls meters in = seconds The total travel time in seconds is = = = + + = (c) It will help to make a chart of the time for each descent and each rebound of the ball, together with the velocity just before and just after each bounce Recall that the time in seconds needed to fall meters is The ball hits the ground with velocity = (taking the upward direction to be positive) and rebounds with velocity =,takingtime to reach the top of its bounce, where its velocity is 0 At that point, its height is All these results follow from the formulas for vertical motion with gravitational acceleration : = = = 0 = number of descent time of speed before speed after time of peak descent bounce bounce ascent height The total travel time in seconds is = = +( ) + = + = Another method: We could use part (b) At the top of the bounce, the height is =,so = and the result follows from part (b) c 06 Cengage Learning All Rights Reserved May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part

30 986 CHAPTER INFINITE SEQUENCES AND SERIES 75 ( + ) is a geometric series with =(+) and =(+), so the series converges when = ( + ) + + or + 0 or We calculate the sum of the series and set it equal to : ( + ) = ( + ) = + + =(+) ( + ) 76 + =0 = ± 4 = ± However, the negative root is inadmissible because 0 So = = ( ) is a geometric series with =( ) 0 =and = If,ithassum,so =0 =0 =0 0 = = 9 0 =ln = = ( + ) + = 4 + = [ +] Thus, +and = Since{ } is increasing, =, implying that the harmonic series is divergent 78 Theareabetween = and = for 0 is ( ) = + = ( +) = = ( +) ( +) We can see from the diagram that as, the sum of the areas between the successive curves approaches the area of the unit square, that is, So = ( +) = 79 Let be the diameter of We draw lines from the centers of the to the center of (or ), and using the Pythagorean Theorem, we can write + = + = + = [difference of squares] = 0 Similarly, = + = + =( )( + ) c 06 Cengage Learning All Rights Reserved May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part

31 = ( ) =, = + = + = = SECTION SERIES 987 [ ( + )], and in general, ( + ) = If we actually calculate and from the formulas above, we find that they are 6 = and = 4 respectively, so we suspect that in general, = To prove this, we use induction: Assume that for all ( +), = ( +) = + Then = = + = [telescoping sum] Substituting this into our + + ( +) formula for +,weget + = = =, and the induction is complete + ( +)( +) + + Now, we observe that the partial sums = = = = =, which is what we wanted to prove ( +) of the diameters of the circles approach as ;thatis, 80 = sin, = sin = sin, = sin = sin, Therefore, = sin sin = since this is a geometric series with =sin = sin and sin because 0 8 The series diverges (geometric series with = ) so we cannot say that 0= If is convergent, then =0by Theorem 6, so = Divergence 6=0,andso = is divergent by the Test for 8 = = = = = = =, which exists by hypothesis 84 If were convergent, then ()( )= would be also, by Theorem 8(i) But this is not the case, so must diverge 85 Suppose on the contrary that ( + ) converges Then ( + ) and are convergent series So by Theorem 8(iii), [( + ) ] would also be convergent But [( + ) ]=, a contradiction, since is given to be divergent 86 No For example, take = and = ( ), which both diverge, yet ( + )= 0, which converges with sum 0 c 06 Cengage Learning All Rights Reserved May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part

32 988 CHAPTER INFINITE SEQUENCES AND SERIES 87 The partial sums { } form an increasing sequence, since = 0 for all Also, the sequence { } is bounded since 000 for all So by the Monotonic Sequence Theorem, the sequence of partial sums converges, that is, the series is convergent 88 (a) RHS = (b) (c) = = + = + = + = ( + ) = = LHS = [from part (a)] = = 0= + = because as = [as above] + = + = = =+ 0 0= because as + 89 (a) At the first step, only the interval (length ) is removed At the second step, we remove the intervals 9 9 and , which have a total length of At the third step, we remove intervals, each of length In general, at the nth step we remove intervals, each of length,foralengthof = Thus, the total length of all removed intervals is = + = = geometric series with = and = Notice that at the th step, the leftmost interval that is removed is,soweneverremove0,and0 is in the Cantor set Also, the rightmost interval removed is,sois never removed Some other numbers in the Cantor set are,, 9, 9, 7 9,and 8 9 (b) The area removed at the first step is 9 ;atthesecondstep,8 9 ;atthethirdstep,(8) 9 In general, the area removed at the th step is (8) 9 = 9 8 9, so the total area of all removed squares is = 9 8 = = c 06 Cengage Learning All Rights Reserved May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part

33 SECTION SERIES (a) The its seem to be 5, 8,,, 667,and4 Note that the its appear to be weighted more toward In general, we guess that the it is + (b) + = ( + ) = ( )= = ( ) = = ( ) ( + ) Note that we have used the formula = ( + ) a total of timesinthiscalculation, once for each between and +Nowwecanwrite = +( )+( )+ +( )+( ) = + ( + )= + ( ) = and so = +( ) = +( ) ( ) 9 (a) For = = ( +)!, = =, = + = 5 6, = = 4, = = + ( )= + 4 = = 9 ( +)! The denominators are ( +)!,soaguesswouldbe = 0 ( +)! (b) For =, = =! ( +)!, so the formula holds for = Assume = Then! ( +)! + = = ( +)! ( +)! ( +)! ( +)! + + ( +)! + ( +)! ( +)+ + = + = ( +)! ( +)! ( +)!( +) ( +)! Thus, the formula is true for = + So by induction, the guess is correct (c) ( +)! ( +)! =and so ( +)! = ( +)! = c 06 Cengage Learning All Rights Reserved May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part

34 990 CHAPTER INFINITE SEQUENCES AND SERIES 9 Let = radius of the large circle, = radius of next circle, and so on From the figure we have =60 and cos 60 =,so = and = Therefore, = + + = In general, we have + =, so the total area is = = = + 9 = Since the sides of the triangle have length, = and tan 0 = tan 0 Thus, = = = 6, so = = + = The area of the triangle is, so the circles occupy about 8% 96 4 of the area of the triangle The Integral Test and Estimates of Sums The picture shows that = =, and so on, so =, The integral converges by (78) with =, so the series converges From the first figure, we see that 6 () 5 6 From the second figure, we see that 6 () Thus, we have 6 6 () 5 = = = = The function () = is continuous, positive, and decreasing on [ ), so the Integral Test applies + = Since this improper integral is convergent, the series = is also convergent by the Integral Test c 06 Cengage Learning All Rights Reserved May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part

35 SECTION THE INTEGRAL TEST AND ESTIMATES OF SUMS 99 4 The function () = 0 is continuous, positive, and decreasing on [ ), so the Integral Test applies = 07 Since this improper integral is divergent, the series 0 is also divergent by the Integral Test = 5 The function () = is continuous, positive, and decreasing on [ ), so the Integral Test applies 5 ln(5 ) ln(5 ) 5 ln 4 = Since this improper integral is divergent, the series = is also divergent by the Integral Test 5 6 The function () = is continuous, positive, and decreasing on [ ), so the Integral Test applies ( ) 4 ( ) 4 ( ) 4 Since this improper integral is convergent, the series = ( ) ( ) 9( ) + = 9 7 is also convergent by the Integral Test ( ) 4 7 The function () = is continuous, positive, and decreasing on [ ), so the Integral Test applies ln( +) = [ln( +) ln ] = Since this improper integral is divergent, the series = is also divergent by the Integral Test + 8 The function () = is continuous, positive, and decreasing ()on[ ), so the Integral Test applies = = 0 = 9 Since this improper integral is convergent, the series is also convergent by the Integral Test (): 0 () = ( )+ () = ( +)= ( ) 0 for = = is a -series with =, so it converges by () = = = is a -series with =09999, so it diverges by () The fact that the series begins with =is irrelevant when determining convergence = = Thisisa-series with =, so it converges by () c 06 Cengage Learning All Rights Reserved May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part

36 99 CHAPTER INFINITE SEQUENCES AND SERIES = sotheintegraltestapplies = + + diverges = The function () = is continuous, positive, and decreasing on [ ), ln( +) ln( +) ln 5 =, sotheseries = The function () = is continuous, positive, and decreasing on = 4 4 [ ), so the Integral Test applies 4 4 ln(4 ) ln(4 ) ln =, sotheseries diverges = = = = = Thisisa-series with =, so it converges by () +4 5 = = = + 4 = = + 4 = = is a convergent -series with = 4 = =4 is a constant multiple of a convergent -series with =, so it converges The sum of two = convergent series is convergent, so the original series is convergent 6 The function () = is continuous and positive on [ ) + ( + ) 0 () = = + = ( + ) ( + ) 0 for, so is ( + ) decreasing on [ ), and the Integral Test applies + + ln( + substitution ) with =+ ln( + ) ln =, so the series diverges + = 7 The function () = is continuous, positive, and decreasing on [ ),sowecanapplytheintegraltest tan = tan tan = tan Therefore, the series = +4 converges c 06 Cengage Learning All Rights Reserved May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part

37 SECTION THE INTEGRAL TEST AND ESTIMATES OF SUMS 99 8 The function () = is continuous, positive, and decreasing on [ ), so the Integral Test applies arctan( +) ( +) + so the series = + + converges [arctan( +) arctan ] = arctan, 9 The function () = is continuous and positive on [ ), and is also decreasing since () = (4 + 4)( ) (4 ) = 6 ( 4 +4) ( 4 +4) = ( 4 ) 0 for 4 86, so we can use the ( 4 +4) Integral Test on [ ) ln(4 +4) 4 ln(4 +4) ln 0 =, sotheseries diverges, and it follows that diverges as well 4 +4 = = 0 The function () = 4 = + [by partial fractions] is continuous, positive, and decreasing on [ ) since it is the sum of two such functions, so we can apply the Integral Test 4 + ln +ln( ) [ ln +ln( ) ln]= The integral is divergent, so the series 4 is divergent = () = ln is continuous and positive on [ ), and also decreasing since 0 () = +ln 0 for, so we can (ln ) use the Integral Test ln [ln(ln )] [ln(ln ) ln(ln )] =,sotheseries ln diverges The function () = ln is continuous and positive on [ ), and also decreasing since 0 () = () (ln )() ln = = ln 0 for 65, so we can use the Integral Test ( ) 4 on [ ) ln ln ln + ln + ln + H + ln + = by parts with =ln, =( ) = ln +, so the series = ln converges c 06 Cengage Learning All Rights Reserved May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part

38 994 CHAPTER INFINITE SEQUENCES AND SERIES The function () = = is continuous and positive on [ ), and also decreasing since 0 () = = ( ) = 0 for [and () ()], so we can use the Integral Test on [ ) ( ) ( ) + by parts with =, = H + 0+ =, so the series converges = 4 The function () = = is continuous and positive on [ ), and also decreasing since 0 () = = 0 for 07, so we can use the Integral Test on [ ) ( ) converges + =, so the series 5 The function () = + = + [by partial fractions] is continuous, positive and decreasing on [ ), + sotheintegraltestapplies () + + ln +ln( +) + +ln + ln =0+0+ ln The integral converges, so the series = + converges 6 The function () = is positive, continuous, and decreasing on [ ) [Note that () = = 4 0 on [ )] Thus, we can apply the Integral Test ( 4 +) ( 4 +) 4 + so the series = 4 + converges () +( ) tan ( ) = = [tan ( ) tan ] = = The function () = satisfied for the series cos = is neither positive nor decreasing on [ ), so the hypotheses of the Integral Test are not cos c 06 Cengage Learning All Rights Reserved May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part

39 SECTION THE INTEGRAL TEST AND ESTIMATES OF SUMS The function () = cos is not decreasing on [ ), so the hypotheses of the Integral Test are not satisfied for the + series = cos + 9 We have already shown (in Exercise ) that when =the series = diverges, so assume that 6= (ln ) () = (ln ) is continuous and positive on [ ),and 0 () = +ln (ln ) 0 if +,sothat is eventually decreasing and we can use the Integral Test (ln ) (ln ) (ln ) [for 6= ] (ln ) This it exists whenever 0, so the series converges for 0 () = is positive and continuous on [ ) For 0, clearly decreases on [ );andfor0, ln [ln(ln )] it can be verified that is ultimately decreasing Thus, we can apply the Integral Test = ln [ln(ln )] [ln(ln )] [ln(ln )] + ln + [ln(ln )] + [ln(ln )] +, + + [for 6= ] which exists whenever + 0 If =,then ln(ln(ln )) = Therefore, = converges for ln [ln(ln )] Clearly the series cannot converge if, because then ( + ) 6=0 So assume Then () =( + ) is continuous, positive, and eventually decreasing on [ ), and we can use the Integral Test ( + ) ( + ) + + = ( +) [( + ) + + ] This it exists and is finite + 0,sotheseries ( + ) converges whenever If 0, ln = and the series diverges, so assume 0 () =ln is positive and continuous and 0 () 0 = for,so is eventually decreasing and we can use the Integral Test Integration by parts gives ln [( )ln ] ( ) whenever 0 Thus, = (for 6= ) = ln converges ( ) [( )ln ] +, which exists c 06 Cengage Learning All Rights Reserved May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part