hw-9-sec-12.2-solns

hw-9-sec-12.2-solns - 950 U CHAPTER 12 INFINITE SEQUENCES...

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Unformatted text preview: 950 U CHAPTER 12 INFINITE SEQUENCES AND SERIES 12.2 Series 1. (a) A sequence is an ordered list of numbers \Vhereas‘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. 00 Q) 2 an : 5 means that by adding sufficiently many terms of the series we can get as close as we like to the number 5. n=1 )1 In other words, it means that linin—3c sn 2 5. where 5.»; is the nth partial sum, that is, 3. ~2.40000 ~1.92000 —2.01600 —1.99680 —2.00064 — 1.99987 —2.00003 —1.99999 “2'00000 series witha = —2.4 and r 2 —§, so its sum is Z i 13 n = : = -2. —2.00000 1.1:; r—a) 1 —4 (~ 3) 1. Note that the dot corresponding to n = 1 is part of both {an} and {SN TI-86 Note: To graph {an} and {371 }. set your calculator to Param mode and DrawDot mode. (DrawDot is under GRAPH, MORE, FORMT (F3).) Now under E (t) = make the assignments: xtl=t , ytl=12l (- 5) “t , xt2=t , yt2=sum seq (ytl , t , 1 , t, l) . (sum and seq are under LIST. OPS (F5). MORE.) Under WIND use 1 , 10 , l , 0 , 10 , l , — 3 , l , l to obtain a graph similar to the one above. Then use TRACE (F4) to see the values. 4. 0.50000 1.90000 3.60000 5.42353 7.30814 9.22706 11.16706 13.12091 15.08432 17.05462 2722—1 diverges, since its terms do not approach 0. n2 + 1 ©OOKIC>OTIJ>DJMH 30 The series 2 n = 1 l... O 951.12.: 952 U CHAPTER 12 INFINITE SEQUENCES AND SERIES 8. 0.5 012500 i 1 i 5 0mw7 ; {J .. .{ 0.23333 . ‘ E 026190 i ' ; 0.28274 1 ' , t ' . {an} : 0.29861 - . . , O ‘ 12 0.31111 032121 From the graph and the table, it seems that the series converges. 0.32955 _ 1 1 f" 2 1 52 1 1 1 033654 ______. : _L_.~._;__ = _ _ _ . n(n+2) n 5142 2(12 n+2)'50 f 1 -1 1-1 l/l_l‘1l 1_l .Hsl l-_i_ nflnm+ei“2 2 4 213 5)’2 4 6 2 k k+2 2 _ 1 1 1 1 1 AfibkiziL//,,. “2(2 3 k+1 A+2)' g// As I: ——> be. this sum approaches — 0} = . , 2n 2 . _ _ 9. (a) 11m an 2 hm = —. so the sequence {(1.2} is convergent by 1 12.1.1). 714°C 11-400 3n + 1 3 3C (b) Since lim an = g # 0. the series 2 an is divergent by the Test for Divergence. 71—000 n=l TI. 7). 10. (a) Both 2 a,- and E a j represent the sum of the first 72 terms of the sequence {as that is. the nth partial sum. i=1 j=1 TL Ti. (1)) Z aj = aj + aj + -~+aj : naj,which. in general. is notthe same as at = (11 +a2 + - - - +an. i=1 h—n-‘F—d i=1 nterms @3 + 2 + g + g- + - ~ - is a geometric series with first term a = 3 and common ratio r z Since M = g < 1, the series a __ 3 _i_ converges to 1_r —~ 14/3 —— U3 .— 9. ® g ‘ i + % — 1 + ‘ - - is a geometric series with ratio 7‘ = :2. Since §r§ = 2 > 1, the series diverges. 13. 3 — 4 + -- — — + - - - is a geometric series with ratio r 2 —§. Since Ir} = g > 1, the series diverges. 14. 1 + 0.4 + 0.16 + 0.064 + - -- is a geometric series with ratio r = 0.4 = Since Er} :- CHIN < 1. the series converges to = 60. H 11 1—1‘ 1—0.9 0.1 0° 10n 0° 1010 "‘1 Elven-1:2 ( ) ” SECNON 12.2 SERIES Cl 953 w 10 11—1 . . . . _ . 10 “:1 7:97)": —. 101;1(~—-g) . The latter series 15 geometric With a — 10 and ratio r = —3. Since M = 3539 > 1, the series diverges. o<»(—3)"'1 10° 3"—1 .. . . . 3. 3 . ‘1 T = Z 21 —Z . The latter series 15 geometric wrtha = 1 and ranor = *2. Since Ir! = Z < 1, it conver es to —————1——-—-—— = 5. Thus, the given series converges to l 3— : l. g 1 __ (__3/4) 7 4 7 7 E 1 is a geometric series with ratio 7‘ — —1— Since lrl — i < 1 the series converges Its sum is "=0 wt)" «5' vi ’ ‘ 1 v’i \/§ x/i + 1 f _____.—_—___:__.—-——-—= 2 fi+1 2+\/§. 1—1/\/§ x/ifll «5-1 fi+1 ( ) "go 3: +1 - E1; “2:20 is a geometric series with ratio -r : Zr— 3. Since [r] > 1, the series diverges. °° e" °° e 23 3H “ 3 Z (' 3)” is a geometric series with first term 3(e / 3) = e and ratio r = E. Since [1-] < 1, the series n=1 3 . e 3e . It —-—-———— = . converges 5 sum 15 1 _ 6/3 3 _ e co1_1c><>1d. . hf. .1 .1. h d. .1 H .. E1 5; _. 5 "2:31 5 iverges smce eac 0 its partia sums 15 5 timest e correspon mg partia sum 0 t e harmonic series °° 1 . . Z -, which diverges. n=1 n x 1 °° 1 If 2 g were to converge, then X E would also have to converge by Theorem 80).] 71:1 r n=1 In general, constant multiples of divergent series are divergent. 22. :1 27:11:13 diverges by the Test for Divergence since lim an . n + 1 1 n—x _nh—Kale2n——3—§#O’ cckzd' thfD' ‘ 1i 1' H 23. E32 k2 _ 1 merges yt e est or ivergence Since “incur. —— k2 _ 1 _ 1 7i 0. kg} w diverges by the Test for Divergence since klin; at, = klirr; = [Clint-:0 = 1 ¢ 0. 25. Converges. oc- 1’1 x n x 1 n n 2 1 + 2 Z —1' + 2- : —- + 3 [sum of two convergent geometric series] n21 3” n=1 3” 3" 71:1 3 3 = .._1’/_3_ + _._2_L§__. _—_ l + 2 : § 1 — 1/3 1 — 2/3 2 2 w1+3‘" 001‘3"_c’c 1",3"_°=1”°°3" .. 26. El 2n —— "2:21 (-2—; -r 571-) —— “Zr-:1 x 2 — E1 2 + ngl 5 . The first series is aconvergent geometric series (fr! = i— < 1), but the second series is a divergent geometric series (Ir! 2 3;- Z 1), so the original series is divergent. 954 28. Sec, i224 CI CHAPTER 12 INFINITE SEQUENCES AND SERIES 30 x 2 (0.8)”‘1 — 2 (0.3)" [difference of two convergent geometric series] f {(0.8)"‘1 — (0.3)"] — « 11:1 71:1 11:1 1 0.3 g ______.._nv- fl “1—0.8 1—0.3‘” 7 diverges by the Test for Divergence since 2; 2_;_ lim an = limln<-n—'——1-)=In<lim 2mg ¢ 0. vii—~00 nr—vx- 2712 + 1 n—~x 2712 —i- 1 00 . 2 (cos 1)""‘ is a geometric series with ratio r = cos 1 % 0.540302. lt converges because Ir! < 1. Its sum is k=1 cos 1 ——-——z .17~ 43. 1~cosl 1 D3 00 . Z arctan n diverges by the Test for Divergence since lirn an : lim arctann = gé 0. 3'4. 35. “:1 n—vx n-~oc 1 2 0C 00 3C 3: . ( 3 + diverges because 2 g- : 2 Z i— diverges. (If it converged. then —- v 2 E i- would also converge by 1 71:1 7%: 5n 71 71:1 n n=1 Theorem 8(i), but we know from Example 7 that the harmonic series -1- diverges.) If the given series converges, then the n=1 I 00 00 x» 00 difference 2 + — Z 53; must converge (since Z —3— is a convergent geometric series) and equal 2 2—, but we : n=1 n 72:1 5” 11:1 n °° 2 . . . have just seen that Z Z diverges, so the given series must also diverge. n=1 co 1 0° 1 "' . i . . 1 . 1 , 1 . . Z —n z E — is a geometric series wrth first term a : — and ratio r = ——. Since Ir] 2 —- < 1, the series converges 6 11:1 8 “=1 6' e C - E = 1 By Example 6, 1 1/6 1/6 e—l' n=1n(n+1) to1~1/e:1-1/e 00e" 6 en en 2 —2 diverges by the Test for Divergence since lim 0Lfl : lim 7 = lim 7 = lim — = n n—>oo n—oc ’n x~>oo :1: .‘c—aoo 21: = 1. Thus, by Theorem 8(ii), 11:1 ()0 Using partial fractions, the partial sums of the series 2 n 2 s”=.§2 (i—1)(2’+1):~ ruse-g 956 U CHAPTER 12 lNFINlTE SEQUENCES AND SERIES —— 417 417 417 417 . 43. .4 = —— ~— '—-— ———— tri 'th ~- . 3 17 3+ 103 106 + Nov» 103 + 106 + isageome cseriesw1 103 an 103 a 417/103 417/103 417 .— - It I converges 0 1 _ T 1 _ 1/103 999/103 999 us, 7 9 9 99 33 54/103 62 54 6192 344 .— 54 54 44. . 4:. —— —— ---=.2+_——————=—— ——=——_ 625 62+103+105+ 6 14/102 10+990 990 .55 __ 42 42 42 42 42 1 4' _ :- ____ y____ . ' ' :~—— :—-—-, 5 1 5342 1 53+ 104 106 Now 104 + 106 -r- isageomemc seneswitha 104 andr 102 Itconver esto a g 1 — 'r ' 1 — 1/102 ’ 99/102 " 9900' ._ 42 153 42 15 147 42 15 189 5063 . =1. ————:——+—-——=—-——-‘ -—-= ’ -—-’- Thus’ 1 5342 53 + 9900 100 9900 9900 + 9900 9900 °r 3300 12’345 ——————-12’345 + - -- Now ——————12’345 J— —————-12‘345 + - -- isa eometrie series witha - 12’345 and — l- ‘ ’ 1010 g " 105 ‘ 105' 46. 7.12345 = 7 + ~165— + 1010 105 _ 12,345/105 * 12,345/105 _ 12,345 ” 1 — 1/105 ” 99,999/105 ’ 99,999 12 345 699 993 12 345 712 333 237 446 . 2 = __:__.. __._:_~ ___’__ = _.L_ ___’__.. Thu“ 1 5 7+ 99,999 99,999 + 99,999 99,999 m 33,333 a 1—1‘ It converges to °° :c °° a: n . . . . x . 1x] 47. 2 ~— 2 2 isageometric series wrthr = —, so the series converges <=> M < 1 <=> ~3— < 1 42> M < 3; n=13n 71:1 s,—-3 <51: <3.lnthatcase,thesumoftheseriesis lir T—i—gfi = lf/i/g - g = 3fx. that i 00 Z (:c — 4)" is a geometric series with r = x — n21 . . 13—4 17—4 3 < cc < 5. In that case, the sum of the series is ——-—-——-—- :: . 1—(22-4) 5—1: 4, so the series converges <=> M < 1 <=> lac —— 41 < 1 «e: ith r = 436, so the series converges <=> M < 1 4=> 41:01 < 1 ¢:> co 00 )3 4“:v"’ = 2 (4:2)" is a geometric seriesw —o 11:0 1;. M < i. In that case, the sum of the series is 1 __1 4x. 50. Z (w :7?) is ageometric series with r = m g 3, so the series converges <=> M < 1 <23» 1:833! < 1 '4: 11:0 \m + 3} < 2 <=> —5 < ac < —1. Forthese values ofz, the sumofthe series is -—————l-———- = fl—2——- = — 2 1~—(a:+3)/2 2—(zc+3) x+1 oo 71 1 51. 2 co; m is a geometric series with first term 1 and ratio r = CO: CD , so it converges ¢> M < 1. But M = ‘60:“ 5 —2— 11:0 for all 9:. Thus, the series converges for all real values of a: and the sum of the series is —-—-——1-—-- = 2 . 1 —— (coscc)/2 2 — cosa: 1 a 0 and in is continuous, we have lim 111(1 + = ln 1 = 0. ".406 52. Because —— n (11:1) = Z [1n(n + 1) — in n] diverges. 71:1 We now show that the series 2 111(1 + = Z In 71:1 ' 11:1 sn =(ln2—1n1)+(ln3—1112)+-~+(ln(n+1)——1nn)=1n(n+1)——ln1=ln(n+1). = ln(n + 1) —+ 00, so the series diverges. As n —+ 00, 5n 952 D CHAPTER 12 lNFiNIT E SEQUENCES AND SERiES 0.5 2 0.12500 i n- i {5’5} - 3 0.1916: g . , . - f 4 0.23333 . ‘ S 5 0.26190 ’ ' g 6 0.28274 ' g 1 v {ca} i 7 0.29861 1 ' - . . , , an 8 0.31111 0 9 (132121 From the graph and the table, it seems that the series converges. 10 0.32955 ‘2 ‘ 1 i=2 1-" 1 1 1 11 .33654 .___=_r_____;...=_ _~ .- 0 n(_n+2) n n~2 2(72 n+2) 30 f; 1 -1 ii.“ 12-3 +£1-11...1l l___1_. n=2n(n42}-2 2 4) 2 3 5 2 4 6 2 k k+2 _ l 3 _ l _ __1._ _ __1_) “ 2 2 3 1 — 1 k 4- 2 ‘ As Is A x, this sum approaches — 0) = . . 2n 2 1 . 9. (a) hm an = hm = —. so the sequence {as} 15 convergent bytllil ). n-wc "doc 3n + 1 3 x (b) Since Iim (2.,I = % 51$ 0. the series Z an is divergent by the Test {Or Divergence. n—roe “:1 21 1'1 10. (a) Both Z 41,-, and Z aj represent the sum of the first 1? terms of the sequence {on }. that is. the nth partial sum. i=1 i=1 H Ti (b) Zaj =aj+aj +--- +0,- == naj.whiciLingeneraLisnotthesameas 201 :01 +a2 +--~+a,,. i=1 W—u—t i=1 nterms @3+ 2 + 4; + ~3- + - ~ - is ageometric series with first term a = Sand common ratio r = Since i-ri = g < 1. the series «1 ~.__3 ——__-°.».. .. convergestohr ._ 14/3 -— 1/3 —9. 63% " i + i- — 1 + - w is a geometric series with ratio r = ~12. Since 2r: 2 2 > 1, the series diverges. 13. 3 - 4 + 133‘- - % + ~ - - is a geometric series with ratio r = —§. Since {r} := g > 1, the series diverges. I 14. 1 + 0.4 + 0.16 + 0.064 + m is a geometric series with ratio r = 0.4 = Since in 2 who < 1, the series converges to @ 6(0.!~))"‘1 is a geometric series with first term a = 6 and ratio r = 0.9. Since {7*} = 0.9 < 1, the series converges to n=1 0 6 6 ...
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hw-9-sec-12.2-solns - 950 U CHAPTER 12 INFINITE SEQUENCES...

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