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hw-4-sec-12.6-solns - V6 61 11‘ ly 1e series this...

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Unformatted text preview: V6. 61'. 11‘ ly 1e series ; this mating :hQT‘hla true asa SECTION 12.6 , ABSOLUTE CONVERGENCE AND THE RATIO AND ROOT TESTS 981 12.6 Absolute Convergence and the Ratio and Root Tests 1. (21) Since lirn an“ : 8 > 1, part (b) of the Ratio Test tells us that the series 2 an is divergent. 71"“) an (b Since 11m an“ : 0.8 < 1, part a of the Ratio Test tells us that the series an is absolutely conver ent and n—*oo a g TL therefore convergent). (0) Since lirn an“ : 1, the Ratio Test fails and the series 2 an might converge or it might diverge. 71—100 0," 2 2 n 2 1 2 1 1 @l‘he series ”:1 g— has positive terms and "lingo (12:1 = ”11,120 [Lg—2% - E] = ”1320(1 + E) . é : 5 < 1, so the series is absolutely convergent by the Ratio Test. —10 n+1 (—1o)"+1' n! (n+1)! (—1o)n : lirn 714.00 } = 0 < 1, so the series is 71-400 @220 (_:L?) . Using the Ratio Test, lirn absolutely convergent. 2TL 2” 271 Vim-1)” 1— diverges by the Test for Divergence 11111 ~— A —00, so lim (—1)" 1 does not exist. n—>oo n4 n—>oo ”—4 °° <—1>"+1 6/77 is conditionally convergent. 6) is a divergent p—series (p = i S 1), so the given series n=1 converges by the Alternating Series Test, but tZ‘Bn §| )n is absolutely convergent. EMS 021—4 is a convergent p- -series (p: 4 > 1) n _. (k+1)(§)k+1_. k+121_2 1 _2 _2 . fkl:ngo[ k(§)k _kl:n;o—k— 3 316131010 1+— ~§(1)#§<1,sotheser1es OO "12 l~c(%)lC is absolutely convergent by the Ratio Test. Since the terms of this series are positive, absolute convergence is the same as convergence. . n 1 n 1 l 8 ”lingo do: _ "1320 [(110623 . $3.]: "Ii—{1:0 ”13.0 g —00, so the series 21180” diverges by the Ratio Test. n+1 4 1. 4 9 lim an+1 :lim[(i1—)——4 n ]= lim fl———4=(1.1 ) 11m ———1—4-—(1 1) lim ——1——4 n—voo an n—>oo (n + 1) (1 1)" 71—>OO (n + 1) n~—>oo (71 + 1)4 n—>oo (1 + 1/n) n4 (1.1)" so the series 2 (#1)" diverges by the Ratio Test. 71:1 3 \ If: // : 982 D CHAPTER 12 INFINITE SEQUENCES AND SERIES :2 I 5:, L) 3 10 i ( 1)"——n——— converges by the Alternating Series Test (see Exercise 12 5 8) Leta — -1— with b * n if 'm x/n3+2 ' ' ' " W1 " «713+? n . 1 x/ 3 2 . 3 . Then lim a_ : 11m — L : 11m———— 71 +2 : 11m 1 + — :1 > 0, so 2—— diverges by naoo n naoo J; n n—>oo n—voo ”3131/11, +2 limit comparision with the divergent p-series :3: \/_ [p21 _ <1]. Thus, 2 («1)71 is conditionally convergent. ":1 V n3 ‘I' 2 81/11 6 oo 61/" 11. Since 0 <8 n3 g E : 6<F> and :17: — is a convergentp— —series [p— — 3 > 1], 2 7;- converges and so oo ‘ TL l/n ;1( 1): is absolutely convergent. sin 4n 1 0" sin 4n 1 1 5" 12. 4 S 4—7, so 2 4 converges by comparison with the convergent geometric series 2—- 4" [IT‘I : Z < 1]. it n L ”:1 " 71:1 f' 00 ' 4 . .; Thus, 2: Sinnn is absolutely convergent. I‘ 11:1 1‘ 1 , an+1 _ _ 1071+1 (n +1)42"+1 ‘ . 10 n + 1 5 10“ g 6 3320 an — 7.1520 (n+2)42n+3 ' 1071 “1320 1'5 ' n+2 : E < 1 ”thesenes 2 (71+ 1)42n+1 is absolutely convergent b the Ratio Test. Since the terms of this series are positive, absolute convergence is the same as I. y 5%. convergence. .; " 71 2 2 n , an+1 i , (n+1)22+1 n! _ . 1 2 _ . 0° Mn:I 33 (11 .132. an — .131;on ‘ n22n — $32. 1 + a ‘ n + 1 r 0’ ”themes .25 1)" S absolutely convergent by the Ratio Test. —1 n 2 0° 2 0° 1 0° —1 " 15. D—gm <——— 712— so since ”:1 Kin/T : £7121 n— converges (p— — 2 > 1), the given series g (—lgw converges absolutely by the Comparison Test. 16 712/3 —2 > Oforn > 3 so m > —1— > 1 form > 3 Since 2 diverges [p2 3 < 1] sodoes ' — ’ ”2/3 _ 2 ”2/3 _ 2 712/3 ”2/3 3 — ’ 00 # z §2—/éC—Os—g by the Comparison Test. 71:1 77» _ 0° (—1)" . . . . 1 1 . . 17. Z converges by the Alternatlng Series Test s1nce 11m — = 0 and —— 1s decreasmg. Now lnn < n, so ":2 1n n - n—>oo nn Inn 1 >1 1 —— —— ,and since 2'1 is the divergent (partial) harmonic series, 2 —— diverges by the Comparison Test. Thus, lnn> n n_ 2 n n_21nn 2| (1111"; is conditionally convergent. I 71:2 n 1 1 1 "+1 . n 1 1 n! 18.11m (1+1 — lim (11+ )/(n+) — 11m ——n——,;= lim———n=—<1,sotheseries§:oO n—wo 0,7L n—wo nI/n" n—eoo (n -I— 1) 71400 (1 —I— 1/n) 6 n=1 n” converges absolutely by the Ratio Test. I 1 I I SECTION 12.6 ABSOLUTE CONVERGENCE AND THE RATIO AND ROOT TESTS 19. |cos (n7r/3)|< 0° cos(mr/3) w fiand "21;:— converges (use the Ratio Test) so the series 7112 converges absolutely by the n! Comparison Test. 2 00 — lim — — —0 < 1, so the series Z 71:1 11—400 T}, (-2)"_ nn (-2)” nn 20, lim V/lanl: lim " n—»oo ”“00 is absolutely convergent by the Root Test. .. n2+1 . 1——1/n2 1 . 0° n2+1 ”. "lingo \/ |an|— - "lingo 2712 + 1211131010 m = E < 1, so the ser1es ”2:31 2712 + 1 IS absolutely convergent by the Root Test. 25715 1 1 2_ 1' 7W n : 11111 n = lim — :3211m ——5 2321 2 ”31:0 ia i n—too n—roo (TL + 1)5 n—>oo 71 + 1 nl—vngo(1 + 1/”)5 n 2 32(1) 2 32 > 1, —2n n+1 571 so the series ”2 2< > diverges by the Root Test. Race 2 n n . 1 n . . Q lim \"/|an| : lirn (1 + i) = 11111 (1 + E) 2 e > 1 [by Equat1on 7.4.9 (or 7.4*.9)], so the serles 2 Z (1 + 1) diverges by the Root Test. ”711/ L/n lirn " |a\/—n|— — —nlin;Q " Uni)" — "151507111171 (30 < 1, so the series ”A: (lnn)“ is absolutely convergent by the Root Test. 1/3. 1 11196 1-1 . 1/20 . 0 (at) Letyzx .Thenlnyz mlnm, so lim lny: lim 7 2 11111 ~— :0 => 11m y=e =1. 25. Use the Ratio Test with the series . 1- 3 1~3~5 1-3-5-7 n_11‘3~5~--~-(2n—1) 0° _11 35 ~(2n—1) _ __ v __ ... _1 __—_ = _1 ___—_. 1 3! + 51 7! + H ) (2n — 1)! + E} ) (2n— 1)! , an“ _ , (~1)”-1~3-5 ----- (2n—1)[2(n+1)—1]. (271—1)! n15}; an “ 7.1330 [2(n + 1) w 1]! (—1)n-1.1-3-5 ----- (271 ~ 1) (—1)(2n+ 1)(2n7 1)! 1 : l — w l — =0 1, ”in; (271 + 1)(2n)(2n — 917131010 271 < so the given series is absolutely convergent and therefore convergent. 2-6 246410 26 10 14+ °° --6 10 14- -(4n—2) 6. th th — ~— —— ————————. 2 UsetheRatioTestwi eserie55+5 8+5~8-11+5 8 11 14+=n1§5 8 11 14 ~(3n+2) , an“ . 2 6 10 ----- (4n—2)[4(n+1)~2] 5-8-11 ----- (3714—2) . 4n~~2 4 .3320 an 7.13;, 5 8 11 ..... (3n+2)[3(n+1)+2] 2-6-10 ----- (4n—2) 7.320 3n—5 3 > ’ so the g1ven series 1s dlvergent 00 ..... oo . ..... 00 n I oo 27. Z 2 4 6 I (271) Z (2 1) (2 2) (2' 3) (2 n) = Z 27171;. = Z 2",Which diverges by the Test for 71:1 TL ’l’L 71:1 ~ n=1 Divergence since lim 2" = 00. n—MX) / .11 r \_ a t. I 5 9 o 984 CHAPTER 12 INFINITE SEQUENCES AND SERIES 2"+1 (n + 1)! n . - 8 - 11 ----- 3 5 . 2 1 . 28 lim (1 +1 : 11m —5——n(—nj—L—) = hm m = E < 1, so the series converges absolutely by the n—yoo an 11-+oo 2 TI! n—NX) 3n + 5 3 5 ~ 8 - 11 ----- (3n + 2) Ratio Test. . . . n . 1 . . . 29. By the recursive defin1t10n, hm a +1 2 hm 5n + : § > 1, so the senes dlverges by the Ratio Test. Wit—+00 an naoo 4n + 3 4 . . . . n . 2 . , 30. By the recurswe defin1tion, hm (la-H : hm 1%? = 0 < 1, so the serles converges absolutely by the Ratlo Test. n—voo n n—»00 n of?" 3 3 » . 1 . iii/(a) hm M — lim _n_3 : lim -——— =1. lnconcluswe . n—>oo 1/71, n—>oo (TL + 1)3 n—Dm (1 + 1/”)3 . (n+1) 2” a. n+1_. 1 1 _1 (b) "113:0 2” +1 ~ : — "ll—{2° 2n — ”lingo 5 + g — —. Conclusive (convergent) . (—3)" fl . c hm ~———— 2 3 hm — 3 lim w —3 Conclusive diver ent ( )n—Nx) 1/71 + 1 (_3)n—1 n—+oo 71+ n—voo +11/n ( g ) t/ 1 2 2 1 (d) lim —"+—2 - 1 “L n = lim 41+ 1 1 —1/—”+—2 = 1. Inconclusive woo 1+(n+1) W woo n 1/n2+(1+1/n) 32. We use the Ratio Test: hm an“ 1 hm [(71 + DIP/[Mn +111! _ hm (n + 112 (n + 1)2 Now if k = 1, then this is equal to lim 11—»00 : 00, so the series diverges; if k 2 2, the limit is (n+1) lim mm) ‘m — —— < 1, so the series converges, and if k > 2, then the highest power of n in the denominator is (n+1)2 ,_: larger than 2, and so the limit is 0, indicating convergence. So the series converges for k 2 2. n+1 I 1 . 33. (a) 11m ”n+1 : lim ”7" 1 : lim ”5 = m lim 2 m . 0 = 0 < 1, so by the Ratlo Test the n—>oo an Ham (7], + 1)! a)” n—>oo n + l naoo n + 1 00 1:71 series I converges for all 2:. n=0 TL. (b) Since the series of part (a) always converges we must have lim %— — 0 by Theorem 12. 2. 6. 34- (a) Ru 2 an+1 + an+2 “I" (It—1+3 + an+4 + ‘ " = an+1<1 + an+2 + an+3 + (171.4 + - - -> an+1 an+1 arr—1 an~2 a “3 an+2 an+4 an+3 an+2 :an+1<1+ + " I +--- an~~1 an »2 an+1 an+3 an+2 an+1 : an+1(1 + rn+1 + rn+2rn+1 -- Tn+3Tn+2Tn+1 + ' ’ ) (*) an+1 S an+1 (1 + rn+ + Ti“ + riul + - - -) [since {Tn} is decreasing] : 1 — Tn+1 (b) Note that since {Tn} is increasing and Tn —~> L as n ~+ 00, we have 7",; < L for all n. So, starting with equation (k), a Rn = an+1(1+ Tn+1 + Tn+2Tn+1 + Tn+3Tn+2Tn+1 ‘I' ‘ ' ') S “n+1 (1 ‘I' L + L2 + L3 + ' ' - )‘Z 1 ”:2 ...
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