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hw-4-sec-12.7-solns

# hw-4-sec-12.7-solns - «a SECTION 12.7 STRATEGY FOR TESTING...

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Unformatted text preview: «a SECTION 12.7 STRATEGY FOR TESTING SERIES 987 . i . . 1 1 _ . . 6. Use the Limit Comparison Test With an 2 2n + 1 and b" 2 E: “131010 %3 2 “lingo 2n: 1 2 ﬁlings m 2 g > 0. . . . °° 1 . 0° 1 Since the harmonic series 2 — diverges, so does 2 . [0r: Use the Integral Test] 1121 n n21 2” + 1 e 1 . . . i . 7. Let f 2 l . Then f 18 posmve, continuous, and decreasmg on [2, 00), so we can apply the Integral Test. a: n at u 2 In 3:, Since/ 1 dnc 2/u’1/2du22u1/2+C22Vlncc+C,weﬁnd :10 lnm du 2 dm/m 00 t t / d? 2 lim dry 2 lim [2v1n93] 2 lim (2 vlnt ~ 2 vln 2) 2 00. Since the integral diverges, the 2 :v n x 2 t—too 2 x 111\$ t—voo t—>oo 1 given series 2 diverges. n=2 n Inn 8 co 2%! 00 2k U . h R . T . 2 ———-——. Sl t e atio est, we et .2 (k + 2)! E. (k: + me + 2) “g g , ak+ _ . 2"+1 (k+1)(k+2) _ . k+1 _ . . [91:11:30 ak — klirrio (k + 2)(k + 3) 2k — [\$13.10 2 ——k + 3 — 2 > 1, so the series diverges. 0r: Use the Test for Divergence. oo 00 2 9. Z We” 2 Z Using the Ratio Test, we get 1:21 k21 6 2 . ak+1 . (k+1)2 ek . k+1 1 21 1 . 2 —— . —— : 1 —— - — 2 . _ : ._ ' 1611120 ak [clinic 6H1 W k k e 1 e e < 1, so the series converges 2 _z3 . . .. , 50(2 — 3mg) . 10. Let ﬂat) 2 x e . Then f is continuous and pOSitive on [1, 00), and f 2 —3— < 0 form 2 1, so f is 61 3 t l 21 _ 1 ' 3 6 J — ——38, so the integral decreasing on [1, 00) as well, and we can apply the Integral Test. floo m2e‘ms dm : lim [— 1 taboo converges, and hence, the series converges. 00 _ n+1 11. bn 2 “1:” > O for n 2 2, {bn} is decreasing, and “lingo bn 2 0, so the given series “22:2 (71—1131;— converges by the Alternating Series Test. 00 . 12. The series 2 sin n diverges by the Test for Divergence since lirn sin n does not exist. 7121 71200 n+1 1 2 I 1 2 00 n 2 3. m an“ 2lim 3 (n+) n 2lim§m23lim n+1:0<1,sotheserieszgn 2.00 an 11200 (n + 1)! 371712 new (71 + 1M2 “#00 n2 1121 n! ., converges by the Ratio Test. ' 2 1 1 1 n . °° ' 2 . . . . 14. 181: 27: S 1 + 2” < 2—” 2 < , so the series 712:; :1: 27: converges by comparison With the geometric series sin 2n converges absolutely, implying convergence. 2 with |r| 2 % < 1. Thus, the series Z 77.21 7121 1+2” 988 CHAPTER 12 INFINITE SEQUENCES AND SERIES , an“ _ (n+1)! 2-5-8---~(3n+2) . n+1 1 1 . = l ' - — 1m 2 — < 1, 5 7111—120 an nggoiQ-5~8~~-(3n+2)[3(n+1)+2] n! n1—>oo3n+5 3 oo 1 so the series "2:20 2 . 5 . 8 . . (3n +2) converges by the Ratio Test. if” A . . . . . n2 —— 1 1 16. Sing the Limlt Comparison Test With an 2 n3 __ 1 and b,L : R, we have 2 3 2 77. . . . 1 n 1 . 0° . . . “lingo 2—“ : (:3 - : "1330 77:3 : “111210 ﬁﬁg : 1 > 0. Since 7;} b.L is the divergent harmonic ()0 series, 2 a,L is also divergent. 71:1 17. lim 21/" = 20 = 1, so lim (#1)n 21/" does not exist and the series 2 (—1)"21/" diverges by the Test for Divergence. n—aoo '71—’00 n:1 y/"i 1 . . .. . 0° (~1)"‘1 . 18 n : for n 2 2. {bn} is a decreasmg sequence ofposmve numbers and 11m 1),. 2 0, so Z —— converges by ‘- W — 1 n—wo “:2 n A 1 the Alternating Series Test. 19 Let f(:n) # LIE— Then f’(\$) — 2 7 mm < Owhenlna: > 2orac > 62 so ln—n is decreasin form > e2 . — — 2x3” , J7; g - By l’Hospital’s Rule lim FLT-L ~ lim i- # lim -2— ﬂ 0 so the series :2: (—1)"117£ converges by the i “A” x/ﬁ 71"” 1/ (2 71"“) n , n=1 \/7_i Alternating Series Test. k 6 k 1 . k 6 1 . 0° k 5 . (12:1 2 51:1 - k: 5 = 3 161111010 iii—5 : g < 1, so the series El 5;: converges by the Ratio Test. (—3271 = i 4- n lim "x/|a l = lim £ 2 0 < 1 so the given series is absolutely convergent by the Root Test I n=l 71’” n21 n . "Too 71 “Too 77' g . 22 n2 _ 1 < ——-—n——— < 1L— — i for n > 1 so i ——i2—_—1—- conver es b the Com arison Test with the ln3+2n2+5 n3+2n2+5 n3_n2 —’ nzln3+2n2+5 g y p 00 convergent p-series 2 1/77.2 [p = 2 > 1]. n:1 . . . . . 1 23. Us1ng the Limit Comparison Test With an 2 tan and b" = R, we have 2 I ~ 2 lim (—11 = lim W)- : lim M g km W = lim see2(1/m) ; 12 : 1 > 0, Since n—voo bn n—>oo l/n x—wo 1/(1: x—‘oo ~1/(E2 I400 00 00 2 bn is the divergent harmonic series, 2 an is also divergent. n:1 11:1 24. lim an 2 lim ’71—’00 ’71—’00 1 ' ' . °° . (n sin —) 2 lim M = lim 8mm 2 1 ¢ 0, so the series 2 nsin(1/n) diverges by the 71:1 n n—>oo 1/n xH0+ 9E Test for Divergence. ...
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