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hw-2-sec 12.2-solns

# hw-2-sec 12.2-solns - ’ff""" VJ i 53:x 3 =1 1...

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Unformatted text preview: ’ff'." " " VJ i 53 :x 3 =1 1. E: E ; 952 CHAPTER 12 lNFlNlTE SEQUENCES AND SERIES 0.12500 0.19167 0.23333 0.26190 0.28274 0.29861 0.31111 0.32121 0.32955 0.33654 From the graph and the table, it seems that the series converges. As k H 00, this sum approaches ﬁg - 0) = 155 . . 2n 9. (a) 11m an _. ”11—120 3n+1 TL—‘iOO , so the sequence {an} is convergent by (12.1.1). ODIN) 00 (b) Since hm an 2 % 7E 0, the series 2 an is divergent by the Test for Divergence. n—>oo ":1 n E aj represent the sum of the ﬁrst n terms of the sequence {an}, that is, the nth partial sum. 2:1 10. (a) Both 2 a, and i=1 71 which, in general, is not the same as Z a,- = a1 + a2 + - ‘ ~ + an. (‘3) :01 =aj+aj+”‘+aj '=naj= i=1 \————v——" i=1 nterms @ 3 + 2 + g + g + - - - is a geometric series with ﬁrst term a = 3 and common ratio r = 32; Since [Tl : % < 1, the series a - _3_ _ _ _§_ - converges to 1_T — 142/3 f 1/3 — 9. @ 313— — i + % — 1 + - - - is a geometric series with ratio r : —2. Since [7“ = 2 > 1, the series diverges. 13. 3 — 4 + if — % + - - - is a geometric series with ratio 7" : ﬁg. Since Ir] : 93; > 1, the series diverges. 14. 1 + 0.4 + 0.16 + 0.064 + - - ~ is a geometric series with ratio 7" : 0.4 = %. Since |r| : g < 1, the series converges to a _‘ 1 _ g 1—r” 1—2/5’ 3' 00 . 2 6(09)“1 is a geometric series with ﬁrst term a = 6 and ratio r = 0.9. Since [Ti = 0.9 < 1, the series converges to 11:1 . SECTION 12.2 SERIES E 953 Do 10" 00 10(10)"‘1 00 n @z \ — z \ — oo n~1 -1- — .- The latter series is eometiic with a = l and ratio 7‘ = 4n 4 4 g n:1 n=1 1 ~§. Since Ir] : 43‘ < 1, it 1 _ (_3/4) = %. Thus, the given series converges to (31%;) = %. °° 1 . . 1 1 ’ 1 \ is a geometric series with ratio r = ~ . Since 7’ = § < 1, the series conver es. lts sum is ‘5 l I ﬂ g 1 _ \/§ _ x/i .\/§+1_ _ m‘mrm m‘ﬁl“§+ll*2+ﬁ 0° 7r 1 0° 7r ”. . . . . 7r . f 1 E W = E Z (3) is a geometric series With ratio 7" 2 -. Since i‘ n20 71:0 3 [T] > 1, the series diverges. . 00 e 00 (9 2 W = 3 Z n . . . . i e . . (g) is a geometric series With ﬁrst term 3(6/ 3) = e and ratio r 2 5 Since [7"] < 1, the series :1 ‘ iverges since each of its partial sums is % times the corresponding partial sum of the harmonic series 00 DO 1 Z —, which diverges. [If E g; were to converge, then 2 Tl n=1 nt. 0° 1 22. Z n + 1 diverges by the Test for Divergence Since lim (1 — lim n + 1 = — ¢ 0 ”=1 2 — 3 n—wo n-wo n —— 3 2 °° k2 k2 23. Z k2 1 diverges by the Test for Divergence Since lim ak : klim m = 1 ;£ 0 [6:2 -—»oo goo _ _. 1-(1+2/k)_ 1mm —k—>oo (k+3)2 “k131i, (1+3/k)2 ‘ 1 750' TL 3) ] [sum of two convergent geometric series] I \ ggf £28 1' 954 CHAPTER 12 INFlNlTE SEQUENCES AND SERIES 27. 2 {1/2 : 2 + x/2 + {’72 + \4/2 + - ‘ ~ diverges by the Test for Divergence since 71:1 11m an: lim {75: lim 21/"=2°'=1;£0. 28. E [(0.8)"71 — (0.3)"] = E (0.8)"_1 — 2 (0.3)" [difference oftwo convergent geometric series] 71:1 11:1 11:1 1 0.3 _ 5 _ g g g ‘ 1 — 0.8 1 , 0.3 7 7 0° 712 + 1 . . . 2 §11n<2n2 + 1) d1verges by the Test for Dlvergence smce . . n2+1 , n2+1 7111—1113001” : rtlirgoln<2n2 +1) :ln(n11—>ngo 27121-1) :1n% #0 00 30. 2 (cos 1))“ is a geometric series with ratio 7' = cos 1 5: 0.540302. It converges because [r1 < 1. Its sum is k:1 cosl 1 — cos 1 m 1175343. 00 31. Z arctan n diverges by the Test for Divergence since lim an : lim arctann : g # 0. 71:1 TLAOC TLHOO 32. Z 7 + — d1verges because 2 — = 2 Z — d1verges. (If it converged, then — . 2 Z '— would also converge by T121 5 TL n=l 71 71:1 n 2 71:1 71 1 Theorem 8(i), but we know from Example 7 that the harmonic series 2 — n=1 TL diverges.) If the given series converges, then the . 0° 3 2 0° 3 . 9° 3 . . . °° difference :2: (57 + E) # 7121 7 must converge (smce 7;] 5—“ IS a convergent geometric series) and equal ”2:31 E’ but we 0° 2 . . . have just seen that Z — diverges, so the glven series must also d1verge. n21 n °° 1 °° 1 n . . . . 1 . 1 . 1 A 33. Z —— : Z — 1s a geometrlc ser1es w1th ﬁrst term a = — and ratio 7' = —. Slnce lrl = — < 1, the ser1es converges ”:1 e” ":1 e e e 6 1/6 1/6 6 1 °° 1 .. t : ~ — = . B E 1 6 —— z 1. b Th 8 o 1 _ 1/6 1 _ 1/6 e e _ 1 y xamp e , ”21 ”(n + 1) Thus, y eorem (11), 0° 1 1 0° 1 0° 1 1 1 8 ~ 1 6 21(en+n(n+1)) Elen+§1n(n+1) e~1+ e—1+e—1 e~1 34 :V: i diver esb the Test for Diver ence since lim (1 ' lim i # lirn i 5 lim i E lirn em — 00 ¢ 0 I ":1 n2 g y g n-—~oo n - naoo 77,2 _ z—aoo {132 — \$400 2\$ — z—voo — . °° 2 35. Using partial fractions, the partial sums of the series 2 n2 1 are n22 _ 871:; m2: (i—l—1_i—11—1> —<1 01: 2M: :>«--+<;:—3-n:1>+<n:2—;> 956 CHAPTER12 INFINITE SEQUENCES AND SERIES * ’ ’3 43. 3.4—1723+%+%+~~.N0w%+%+~~ isageometn'cserieswitha: %andr= 1—3—5. It convergesto 1-:1—7‘ = {41:114/111; : 3513313: ': 217% Thus, 311—7: 3+ 3—19; 2 36%; = 13—]??? 44. 6.2ﬁ:6.2+%+%+m:62+ 15:1/11/0122 :%+§§6 :ZLSOE :35434 45.1.5342:1.53+—1%+%+~~.Now%+%+-~isageometricserieswitha: \$02—4andr: 1—355. (1 42/104 _ 42/104 _ 42 1 t —-—— : ———————,# _,___ﬂ tconverges ° 1 — r 1 — 1/102 99/102 9900 42 153 42 15,147 42 15,189 5063 Th 1. 35:1.53 -—:_— __: __: ___, “5’ 5 + 9900 100 + 9900 9900 + 9900 9900 or 3300 ——— 12 345 12 345 12 345 12 345 . . . 12 345 1 4. . : ’ ’ ..m ’ ’ -n i : ’ :-—— 6 7 12345 7 + 105 + 1010 + Now 105 + 1010 + 18 a geometric series With a 05 and r 105. Mower em a ‘ 12,345/105 _ 12,345/105 _ 12,345 g 1 ~r _ 1 — 1/105 " 99,999/i05 ‘ 99,999 _— 12 345 699 993 12 345 712 338 237 446 T _1 : _’__ : ___’._. .4— : ___’__ ’ . hus’ 7 2345 7 + 99,999 99,999 + 99,999 99,999 or 33,333 0° 90” °° ac n . . . . m , |w| 47. Z :9): : Z (3) isageometric series WlthT = 5’ so the series converges <=> M < 1 4:) ’3 < 1 4:» m < 3; n=1 11,21 that is —3 < m < 3 In that case the sum ofthe series is a — 37/3 * 33/3 3 :0: - m ’ ' ’ 1—r—1—m/3_1—m/3 3—3—54 ()0 <3: 2 (cc — 4)" is a geometric series with 1" = :0 ﬂ 4, so the series converges 4:) M < 1 c) lac — 4| < 1 4:) n21 . . — 4 — 4 3 < at < 5. In that case, the sum ofthe series is -—\$———- : a: . 1 4 (:c 4 4) 5 ~ x f" °° 0° . . . . {—49 Z 4"a:" = E (490)" is a geometric series With 7” : 4w, so the series converges 4:} M < 1 <:> 4 |\$| < 1 4:; n=0 n=0 . . 1 |x| < i. In that case, the sum of the series 18 1 _ 41\$. w n 3 50. 2 933233)‘ is a geometric series With“ 2 m : 3, so the series converges 4:} [rl < 1 4:) Im: 1 < 1 ¢> n=0 lac + 3| < 2 (I) —5 < x < —1 Forthese values ofcc the sum ofthe series is ————1——— — —E——— — — 2 ' ’ 1—(ac+3)/2_2—(m+3)' \$+'1' w n 1 51. 2 CO; at is a geometric series with ﬁrst term 1 and ratio 7” = COMB, so it converges 4:) M < 1. But |r| = 1“):ng _<_ 2 n=0 1 2 for all m. Thus, the series converges for all real values of m and the sum of the series is —————- : ———. 1—(cos:r)/2 2#cos:c . . . 1 52. Because 1 -—> 0 and In is continuous, we have him 1n (1 + —) = ln 1 = 0. TL ’11—'00 TL We now show that the series 2 1n (1 + E) = Z ln<n + 1) = Z [ln(n + 1) — ln n] diverges. ”:1 n21 71:1 n 5n =(ln2—ln1)+(ln3—ln2)+-~+(ln(n+1)*lnn) =ln(n+1)—ln1=ln(n+1). As n —> 00, s” : ln(n + 1) —» 00, so the series diverges. ...
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