does not exist. Examples: { }, {sin } 11, It appears that the sequence is approaching 1 2.

Transcription

1 8 INFINITE SEQUENCES AND SERIES 8. Sequences. (a) A sequence is an ordered list of numbers. It can also be dened 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 sufciently large. (c) The terms become large as becomes large. In fact, we can make as large as we like by taking sufciently large.. (a) From Denition, a convergent sequence is a sequence for which lim exists. Examples: {}, { } (b) A divergent sequence is a sequence for which lim does not exist. Examples: {}, {sin }. The rst six terms of are, 5, 7, 4 9, 5, 6. It appears that the sequence is approaching. lim lim 4. {cos()} 9. The sequence does not appear to have a limit. The values will cycle through the rst six numbers in the sequence never approaching a particular number The denominator of the nth term is the nth positive odd integer, so The denominator of the nth term is the ( )st power of, so. 7. { 7 7}. Each term is larger than the preceding one by 5,so ( ) 5( ) The numerator of the th term is and its denominator is ( ). Including the alternating signs, we get () ( ) Eachtermis times the preceding one, so. 0. {5 5 5 }. The average of 5 and is, so we can think of the sequence as alternately adding and to. Thus, ().. 5 ( 5 ) ( ) 5,so 50 5as. Converges 0. ( ),so as. Converges 0. (0),so lim 0by (7). Converges 65

2 654 CHAPTER 8 INFINITE SEQUENCES AND SERIES 4. ( ),so as since lim and lim ( ). Diverges 5. Because the natural exponential function is continuous at 0, Theorem 5 enables us to write lim lim lim() 0 Converges ,so lim 9 lim by (7) with. 5 Converges 7. If,then lim 8 lim Theorem 5, lim tan 8 tan () ( 8) lim Since tan is continuous at,by 4 lim 8 tan 4. Converges 8. Using the last limit law for sequences and the continuity of the square root function, lim lim 9 lim 9 lim 9 9. Converges 9. () Theorem 4. (),so0 Converges 0. (). Now 0 as,so 0 by the Squeeze Theorem and as, but the terms of the sequence { } alternate in sign, so the sequence 5 converges to and the sequence 4 6 converges to. This shows that the given sequence diverges since its terms don t approach a single real number.. 0 as because and. Converges. cos(). As, 0,socos() cos 0 because cos is continuous. Converges.. Since lim H lim H lim 0, it follows from Theorem that lim 0. Converges 4. as,sosince lim arctan,wehave lim arctan cos [since 0 cos ], so since lim 0, cos Converges 6. cos (). Since as, the given sequence diverges. 7. ln ln,so ln( ) H lim ln lim lim lim lim ln,sobytheorem, lim lim converges to 0 by the Squeeze Theorem.. Convergent

3 SECTION 8. SEQUENCES ( ) ( ) 8,so lim 8 lim 8 lim () 8 0 8by Theorem 5, since the function () is continuous at 0. Convergent 9. ( )! ( )! ( )! ( )()( )! 0 as. ( )() Converges 0. sin. and lim 0,so lim 0by the Squeeze Theorem. Converges. { } diverges since the sequence takes on only two values, 0 and, and never stays arbitrarily close to either one (or any other value) for sufciently large. (ln ). lim H (ln )() ln lim lim H (ln ) lim 0,sobyTheorem, lim 0. Convergent. ln( )ln( )ln ln ln as. Convergent 4. 0! ( ) 7 [for ] 0 as, so by the Squeeze Theorem and Theorem 4, {()!} converges to From the graph, it appears that the sequence converges to. {() } converges to 0 by (7), and hence {() } converges to From the graph, it appears that the sequence converges to a number greater than. lim sin lim sin lim sin lim 0

4 656 CHAPTER 8 INFINITE SEQUENCES AND SERIES 7. From the graph, it appears that the sequence converges to. As, 8 so lim , 8. From the graph, it appears that the sequence converges to as Hence, 5 by the Squeeze Theorem. lim 0 Alternate solution: Let ( 5 ).Then ln ( 5 ) lim ln lim so lim ln 5 5,andso 5 converges to 5. H ln 5 ln 5 ln ln 5 5 lim lim 5 ln5, 5 cos 9. From the graph, it appears that the sequence { } is divergent, since it oscillates between and (approximately). To prove this, suppose that { } converges to. If { } converges to,and lim. But,then cos,so lim does not exist. This contradiction shows that { } diverges. 40. From the graph, it appears that the sequence approaches ( ) () () () () 5 0 as 5 ( ) So by the Squeeze Theorem, () converges to (a) 000(06) 060, 60, 90, 4 648,and 5 8. (b) lim 000 lim (06), so the sequence diverges by (7) with 06.

5 SECTION 8. SEQUENCES (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 The number of catsh increases by 8% per month and is decreased by 00 per month, so 0 8% , 08 00, and so on. Thus, (b) Using the recursive formula with ,weget 500, 508, 55(rounding any portion of a catsh), 4 545, ,and 6 574, which is the number of catsh in the pond after six months. 44. if is an even number if is an odd number When, the rst 40 terms are, 4, 7, 5, 6,, 40, 0, 0, 5, 6, 8, 4,,, 4,,, 4,,, 4,,, 4,,, 4,,, 4,,, 4,,, 4,,, 4. When 5, the rst 40 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. 45. (a), 4 for., 4 4, 4 4, 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, lim and hence, the sequence is convergent. 46. (a) Since lim, theterms approach as becomes large. Because we can make as close to as we wish, will also be close, and so lim. (b), 05, , , , , , , , It appears that lim ; hence, the sequence is convergent. (c) If lim then lim also, so must satisfy ( ) (since hastobenon-negativeifitexists).

6 658 CHAPTER 8 INFINITE SEQUENCES AND SERIES 47. (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).if lim /, then lim and lim,somust satisfy 0 5 [since must be positive]., 48. For,,,, 4, 78,,so ( ) (). lim lim (). Alternate solution: Let lim. (We could show the limit exists by showing that { } is bounded and increasing.) Then must satisfy ( ) since the sequence increases, so. 49. is decreasing since bounded since 0 5 for all. Notethat denes an increasing sequence since for () 4 4, ( ) 5 for each. The sequence is 0 () ( 4)() ( )() ( 4) 7 ( 4) 0. The sequence is bounded since 7 for, and for. 5. The terms of () alternate in sign, so the sequence is not monotonic. The rst ve terms are,,, 4,and5. Since lim lim, the sequence is not bounded. 5. denes an increasing sequence since the function () is increasing for. [0 () 0 for.] The sequence is unbounded since as. (It is, however, bounded below by.) 5. 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 limit. must be less than 8 since { } is decreasing, so (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.

7 SECTION 8. SEQUENCES 659, which is the induction hypothesis. 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 Sequence Theorem, lim exists. (b) If lim,then lim also, so 0 ( )( ) 0 [since can t be negative]. 55.,. 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 lim,then lim also, so must satisfy 0 ± 5. 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 nd the limit, we use the fact that lim lim 0 ± 5.But,sowemusthave (08) ln(08) ln(000000) ln(08) ln(000000) ln(000000) ln(08) 69,so must be at least 6 to satisfy the given inequality. 58. (a) If is continuous, then () lim lim ( ) lim lim by Exercise 46(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 (a) Suppose { } converges to. Then lim lim lim ( ) 0 0or. (b) since.

8 660 CHAPTER 8 INFINITE SEQUENCES AND SERIES (c) By part (b), 0, so lim lim 0 0since. 0, 0, etc. In general, 0, By (7) lim 0if. Here (0 ). (d) Let. We rst show, by induction, that if 0,then and. For 0,wehave ( 0 ) 0 0 since 0. So 0. Now we suppose the assertion is true for, thatis, and.then. And ( ) ( ) 0 because. So ( ) 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 lim. 60., 5, , , , , , 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, soboth{ } and { } are bounded monotonic sequences and therefore convergent by the Monotonic Sequence Theorem. Let lim. Then lim also. We have ( ) ( ) ( ) 4,so 4. Taking limits of both sides, we get 4 4 [since 0]. Thus, lim. Similarly, we nd that lim. Since the even terms approach and the odd terms also approach,it follows that the sequence as a whole approaches,thatis, lim.

9 LABORATORY PROJECT LOGISTIC SEQUENCES 66 LABORATORY PROJECT Logistic Sequences. To write such a program in Maple it is best to calculate all the points rst 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)] t0..0)],t0..0,p0..0.5,stylepoint); In Mathematica, we can use the following program: p[0]/ k.5 p[j_]:k*p[j-]*(-p[j-]) PTable[p[t],{t,0}] ListPlot[P] With 0 and 5: With 0 and 5: INSTRUCTOR Both of these sequences seem to converge the rst to about,the USE second to about ONLY

10 66 CHAPTER 8 INFINITE SEQUENCES AND SERIES With 0 7 and 5: With 0 7 and 5: The limit of the sequence seems to depend on, but not on 0.. With and : It seems that eventually the terms uctuate between two values (about 05 and 08 in this case).

11 LABORATORY PROJECT LOGISTIC SEQUENCES 66. With 0 7 and 4: With 0 7 and 45: From the graphs above, it seems that for between 4 and 5, the terms eventually uctuate between four values. In the graph below, the pattern followed by the terms is Note that even for 4 (as in the rst graph), there are four distinct branches ; even after 000 terms, the rst and third terms in the pattern differ by about 0 9, while the rst and fth terms differ by only 0 0.With and 48:

12 664 CHAPTER 8 INFINITE SEQUENCESA AND SERIES , , , , , 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 eeting 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.

13 SECTION 8. SERIES 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 sufciently many terms of the series we can get as close as we like to the number 5. In other words, it means that lim 5,where is the th partial sum, that is, 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 (5) 4 5 Note that the dot corresponding to is part of both { } and { }. 4 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: xtt, yt/(-5)ˆt, xtt, ytsum 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 The series cos diverges, since its terms do not approach 0.

14 666 CHAPTER 8 INFINITE SEQUENCES AND SERIES The series diverges, since its terms do not approach 0. 4 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 From the graph and the table, it seems that the series converges. so, lim.

15 SECTION 8. SERIES From the graph and the table, it seems that the series converges.,so ( ) ( ) 5 4 As, this sum approaches (a) lim lim,sothesequence { } is convergent by (8..). (b) Since lim 60,theseries is divergent by the Test for Divergence. 0. (a) Both and represent the sum of the rst terms of the sequence { },thatis,theth partial sum. (b) terms, which, in general, is not the same as is a geometric series with ratio 4. Since 4, the series diverges is a geometric series with ratio 4. Since 4, the series converges to is a geometric series with ratio.since, the series converges to (5) is a geometric series with ratio 04 5.Since , the series converges to 5. 6(09) is a geometric series with rst term 6and ratio 09. Since 09, the series converges to

16 668 CHAPTER 8 INFINITE SEQUENCES AND SERIES 6. 0 (9) 0(0) (9) 0 0. The latter series is geometric with 0and ratio Since 0 9, the series diverges is a geometric series with ratio. Since, the series diverges. is a geometric series with ratio.since, the series converges. Its sum is diverges by the Test for Divergence since lim lim 60. ( ) diverges by the Test for Divergence since lim ( ) ( ) lim ( ) lim ( ) 6 0. () diverges by the Test for Divergence since lim lim 6 0. cos diverges by the Test for Divergence since lim lim cos cos Converges. 5 [sum of two convergent geometric series] 4.. The rst series is a convergent geometric series ( ), but the second series is a divergent geometric series ( ), so the original series is divergent diverges by the Test for Divergence since lim lim lim (cos ) is a geometric series with ratio cos It converges because. Itssumis cos 754. cos 7. arctan diverges by the Test for Divergence since lim lim arctan 60.

17 NOT FOR SALE SECTION 8. SERIES 669 (08) (0) (08) (0) [difference of two convergent geometric series] is a geometric series with rst term and ratio.since, the series converges to. By Example 6, ( ). Thus, by Theorem 8(ii), ( ) ( ). 5 diverges because diverges. (If it converged, then would also converge by Theorem 8(i), but we know from Example 7 that the harmonic series difference we have just seen that 5 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. Using partial fractions, the partial sums of the series are ( )( ) 4 5 This sum is a telescoping series and. Thus, lim lim.. For the series 4, [using partial fractions]. The latter sum is [telescoping series] Thus, 4 lim lim 5 6. Converges. For the series Thus, ( ), ( ) [using partial fractions]. The latter sum is [telescoping series] ( ) lim lim. Converges 6

18 670 CHAPTER 8 INFINITE SEQUENCES AND SERIES 4. For the series ln, (ln ln ) (ln ln ) (ln ln 4) [ln ln( )]ln ln( ) ln( ) Thus, lim, so the series is divergent. 5. (a) Many people would guess that, but note that consists of an innite number of 9s. (b) , Itssumis ,thatis,. 09 (c) The number has two decimal representations, and [telescoping series] 9 0, which is a geometric series with 09 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 is a geometric series with 0 0 and 0. It converges to Now is a geometric series with and It converges to Thus, , , or ,45 0 5, Now,45 0 5, is a geometric series with, and 0 5. It converges to, , ,9990 5,45 99,999. Thus, 745 7,45 99, ,99 99,999,45 99,999 7,8 99,999 or 7,446, is a geometric series with, so the series converges ; that is,. In that case, the sum of the series is 0 ( ) is a geometric series with., so the series converges 5. For these values of, the sum of the series is ( ) ( ).

19 4. 0 NOT FOR SALE SECTION 8. SERIES 67 cos is a geometric series with rst term and ratio cos cos,soitconverges. But for all. Thus, the series converges for all real values of and the sum of the series is (cos ) cos. 44. Because 0 and ln is continuous, we have lim ln ln0. 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. 45. After dening, Weuseconvert(f,parfrac); in Maple, Apart in Mathematica, or Expand Rational and Simplify in Derive to nd that the general term is ( ) ( ).Sothenth partial sum is ( ) ( ) ( ) The series converges to lim. This can be conrmed by directly computing the sum using sum(f,..infinity); (in Maple), Sum[f,{n,,Infinity}] (in Mathematica), or Calculus Sum (from to )andsimplify (in Derive). 46. See Exercise 45 for specic CAS commands The series converges to lim For, 0since 0.For, Also, lim lim For 6, Also, Sothenth partial sum is ( ) ( ) ( )() ( ) ( ) ( ) () lim lim because lim H lim 0. ln ( ) ( ) INSTRUCTOR USE ONLY

20 67 CHAPTER 8 INFINITE SEQUENCES AND SERIES 49. (a)aftertherstpillistaken,00 mg of the drug is in the body. After the second pill is taken, 00 mg plus 00(5%) mg remains in the body. After the third pill is taken 00 mg plus 00(5%) mg plus [00(5%)](5%) mg remains in the body. This gives us 00 00(005) 00(005) 055 mg of the drug remaining in the body after the patient takes three pills. (b) Continuing the pattern established in part (a), we get 00 00(005) 00(005) 00(005) mg after pills are taken. By (), this sum is 00( 005 ). 005 (c) The amount of the drug remaining in the body in the long run can be approximated by summing the innite series with 00and 005. Thissumis mg (a) We start with ies being released. After day, we have ies plus a new release of ies. After days, we have ies. After days, we have thelongrun,wehaveageometricserieswith and,soitssumis (b) We want the sum to be 0,000,so0, ( ) [from part (a)] 0,000(0) 000. by (). In 5. (a) The rst 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 ( ) (b) lim ( ) lim lim ( ) by (). since 0 lim 0 [since ] [since ] If 08,then 0 and the multiplier is (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

21 SECTION 8. SERIES 67 (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 : number of descent time of descent speed before bounce speed after bounce time of ascent peak height The total travel time in seconds is ( ) 5. Another method: We could use part (b). At the top of the bounce, the height is,so and the result follows from part (b). ( ) 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 : ( ) ( ) () ( ) 0 ± 4 ±. However, the negative root is inadmissible because 0. So. 54. ( ) is a geometric series with ( ) 0 and.if,ithassum,so ln

22 674 CHAPTER 8 INFINITE SEQUENCESA AND SERIES 55. ( ) 4 [ ] Thus, and lim. Since{ } is increasing, lim, implying that the harmonic series is divergent. 56. Theareabetween and for 0 is ( ) 0 0 ( ) ( ) ( ) 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 ( ). 57. 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]. Similarly, ( )( ) ( ), [ ( )], and in general, ( ). If we actually calculate and from the formulas above, we nd that they are 6 and 4 respectively, so we suspect that in general,, ( ).Then formula for,weget ( ). To prove this, we use induction: Assume that for all ( ) [telescoping sum]. Substituting this into our, and the induction is complete. ( )( ) Now, we observe that the partial sums of the diameters of the circles approach as ;thatis,, which is what we wanted to prove. ( )

23 SECTION 8. SERIES sin, sin sin, sin sin,. Therefore, sin sin sin since this is a geometric series with sin and sin because The series diverges (geometric series with ) so we cannot say that If Divergence. is convergent, then lim 0by Theorem 6, so lim 60,andso is divergent by the Test for 6. 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. 6. No. For example, take and (), which both diverge, yet ( ) 0, which converges with sum 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. 64. (a) RHS (b) (c) [from part (a)] lim lim ( ) LHS because as. [as above] lim lim 00 because as.

24 676 CHAPTER 8 INFINITE SEQUENCES AND SERIES 65. (a) At the rst 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 66. (a). In general, at the nth step we remove intervals, each of length,foralengthof. Thus, the total length of all removed intervals is the th step, the leftmost interval that is removed is the rightmost interval removed is are,, 9, 9, 7 9,and 8 9. geometric series with and. Notice that at,soweneverremove0,and0 is in the Cantor set. Also,,sois never removed. Some other numbers in the Cantor set (b) The area removed at the rst step is ;atthesecondstep,8 ;atthethirdstep,(8). In general, the area removed at the th step is (8) 8, so the total area of all removed squares is The limits seem to be 5, 8,,, 667,and4. Note that the limits appear to be weighted more toward. In general, we guess that the limit is. (b) ( ) ( ) ( ) ( ) ( ) Note that we have used the formula ( ) a total of times in this calculation, once for each between and. Now we can write and so ( )( ) ( )( ) ( ) ( ) lim ( ) ( ) () ( ).

25 67. (a) For ( )!,, 5 6, , The denominators are ( )!,soaguesswouldbe SECTION 8. SERIES 677 ( )!. ( )! (b) For,!, so the formula holds for. Assume! ( )!.Then ( )! ( )! ( )! ( )! ( )! ( )! ( )! ( ) ( )! ( )! ( )!( ) ( )! Thus, the formula is true for. So by induction, the guess is correct. (c) lim ( )! lim lim ( )! and so ( )! ( )!. 68. Let radius of the large circle, radius of next circle, and so on. From the gure we have 60 and cos 60,so and. Therefore,. In general, we have, so the total area is Since the sides of the triangle have length, and tan 0.Thus, tan 0 6, so of the area of the triangle The area of the triangle is, so the circles occupy about 8% 4

26 678 CHAPTER 8 INFINITE SEQUENCES AND SERIES 8. The Integral and Comparison Tests; Estimating Sums. The picture shows that, and so on, so,. The integral converges by (5.0.) with, so the series converges.. From the rst gure, we see that 6 () 5. From the second gure, we see that have (). Thus,we 6 () 5.. (a) We cannot say anything about.if for all and is convergent, then could be convergent or divergent. (See the note after Example.) (b) If for all,then is convergent. [This is part (i) of the Comparison Test.] 4. (a) If for all,then is divergent. [This is part (ii) of the Comparison Test.] (b) We cannot say anything about.if for all and is divergent, then could be convergent or divergent. 5. is a -series with. is a geometric series. By (), the -series is convergent if. Inthiscase,,so are the values for which the series converge. A geometric series converges if,so converges if. 6. The function () 5 is continuous, positive, and decreasing on [ ), so the Integral Test applies. lim lim lim Since this improper integral is convergent, the series is also convergent by the Integral Test The function () 5 5 is continuous, positive, and decreasing on [ ), so the Integral Test applies. 5 lim 5 5 lim lim ,so 5 diverges. 4

27 SECTION 8. THE INTEGRAL AND COMPARISON TESTS; ESTIMATING SUMS The function () 4( 4) is continuous, positive, and decreasing on [ ), so the Integral Test applies ( 4) lim ( 4) lim ( 4) lim 4 5, sotheseries 4diverges. for all, so because it is a p-series with. 4 4 for all,so with (the harmonic series).. The series for if it converged, then converges by comparison with 4 diverges by comparison with is a -series with 085,soitdivergesby().Therefore,theseries 085 would have to converge [by Theorem 8(i) in Section 8.]. 085, which converges, which diverges because it is a p-series must also diverge, and are -series with, so they converge by (). Thus, converges by Theorem 8(i) in Section 8.. It follows from Theorem 8(ii) that the given series ( 4 ) also converges Thisisa-series with, so it converges by ()..Thisisa-series with, so it converges by (). 5. () is continuous and positive on [ ). 0 () ( ) 0 for,so is decreasing on [ ). Thus, the Integral Test applies. lim lim [by parts] lim [ ] since lim lim ( ) H lim ( )0and lim 0. Thus, converges. 6. () is continuous and positive on [ ), and also decreasing since 0 () ( ) 0 for, ( ) so we can use the Integral Test [note that is not decreasing on [ )]. lim ln( ) lim ln( )ln 9,sotheseries diverges, and so does the given series,.

28 680 CHAPTER 8 INFINITE SEQUENCES AND SERIES 7. () ln is continuous and positive on [ ), and also decreasing since 0 () ln 0 for, so we can (ln ) use the Integral Test. lim ln [ln(ln )] lim [ln(ln ) ln(ln )],sotheseries ln diverges ,sotheseries 9 convergesbycomparisonwiththe-series [ ]. Another solution: The function () ( 9)is continuous, positive, and decreasing on [ ), so the Integral Test applies. tan lim 9 lim 9 tan lim tan tan. Since the integral converges, so does the series. cos,sotheseries cos convergesbycomparisonwiththe-series [ ] converges by comparison with, which converges because it is a constant multiple of a convergent -series [ ]. The terms of the given series are positive for, which is good enough.. is positive for and ,so converges by comparison with the convergent 4 4 geometric series for all,so 4 diverges by comparison with the divergent geometric series The function () is continuous, positive, and decreasing on [ ), so the Integral Test applies. diverges. lim lim ln lim (ln( ) 0),sotheseries. The function () is continuous, positive, and decreasing on [ ), so the Integral Test applies. lim lim ln lim (ln( )ln 5), sotheseries diverges.

29 SECTION 8. THE INTEGRAL AND COMPARISON TESTS; ESTIMATING SUMS Use the Limit Comparison Test with 4 and 4 : 6. lim lim 4 4 lim Since the geometric series ( ),so. 4 lim 4 diverges, so does or use the Test for Divergence. lim Alternatively, use the Comparison Test with converges by comparison with the convergent -series 7. (),and converges because it is a constant multiple of the convergent -series 8., so the given series converges by the Comparison Test. sin 0 0 and 0 convergent geometric series. 0 0, so the given series converges by comparison with a constant multiple of a 0 9. Use the Limit Comparison Test with sin and.then and are series with positive terms and lim sin() sin lim lim 0 sin () also diverges. sin() lim 0. Since is the divergent harmonic series, [Note that we could also use l Hospital s Rule to evaluate the limit: H cos() lim lim cos cos0.] 0. If 5 and,then lim so lim 5 lim 5 0, 5 diverges by the Limit Comparison Test with the divergent harmonic series (Note that 0 for 6.). We have already shown (in Exercise 7) 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 ) lim (ln ) (ln ) [for 6 ] lim This limit exists whenever 0,, so the series converges for. (ln )

30 68 CHAPTER 8 INFINITE SEQUENCES AND SERIES. (a) () 4 is positive and continuous and 0 () 4 5 is negative for 0, and so the Integral Test applies (b) 0 lim lim 4 0 (0) 000,sotheerrorisatmost () 0 (0) , soweget 08 with error (c) that is, for. 4.So (0) 5,. (a) () is positive and continuous and 0 () is negative for 0, and so the Integral Test applies (b) 0 lim lim 0, so the error is at most ,soweget 645 (the average of and ) with error 0005 (the maximum of and , rounded up). (c).so 000 if () 5 is positive and continuous and 0 () 5 6 is negative for 0, and so the Integral Test applies. Using (), 5 lim If we take 5,then and So () ( ) 6 is continuous, positive, and decreasing on [ ), so the Integral Test applies. Using (), ( ) 6 lim 0( ) 5. To be correct to ve decimal places, we want 0( ) 5 0( ) 5 ( ) 5 0, ,000,souse ( )

31 6. () NOT FOR SALE SECTION 8. THE INTEGRAL AND COMPARISON TESTS; ESTIMATING SUMS 68 (ln ) is positive and continuous and 0 ln () is negative for, so the Integral Test applies. (ln ) Using (), we need 00 (ln ) lim ln ln. This is true for 00,sowewouldhavetotakethis many terms, which would be problematic because , Now lim lim ,sotheerroris sin sin sin sin sin Nowsin, so the error is lim lim (a) From the gure, (),sowith (), 4 ln. Thus, 4 ln. (b) By part (a), 0 6 ln and 0 9 ln is a convergent -series with 00. Using(),weget lim 000 lim We want , , ,0. 4. Since for each,andsince will always converge by the Comparison Test. 9 0 is a convergent geometric series 0, 0 ln( ) 4. First we observe that, by l Hospital s Rule, lim lim 0 0. Also, if converges, then lim 0by ln( ) ln( ) Theorem Therefore, lim lim 0. We are given that is convergent and 0. 0 Thus, ln( ) is convergent by the Limit Comparison Test. 0

32 684 CHAPTER 8 INFINITE SEQUENCES AND SERIES 4. Yes. Since is a convergent series with positive terms, lim 0by Theorem 8..6, and sin( ) is a sin( ) series with positive terms (for large enough ). We have lim lim 0 by Theorem... Thus, is also convergent by the Limit Comparison Test. 44. ln ln ln ln ln ln.thisisa-series, which converges for all such that ln ln ln [with 0]. 45. lim lim, so we apply the Limit Comparison Test with.since lim 0 we know that either both series converge or both series diverge, and we also know that divergent. 46. For the series Thus,, 4 4 diverges [-series with ]. Therefore, must be () 4 lim lim (). Since a constant multiple of a divergent series is divergent, the last limit exists only if 0, so the original series converges only if. 8.4 Other Convergence Tests. (a) An alternating series is a series whose terms are alternately positive and negative. (b) An alternating series (),where,convergesif0 for all and lim 0. (This is the Alternating Series Test.) (c) The error involved in using the partial sum as an approximation to the total sum is the remainder and the size of the error is smaller than ;thatis,. (This is the Alternating Series Estimation Theorem.). (a) Since lim (b) Since lim 8, part (b) of the Ratio Test tells us that the series is divergent. 08, part (a) of the Ratio Test tells us that the series is absolutely convergent (and therefore convergent). (c) Since lim, the Ratio Test fails and the series might converge or it might diverge.