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hw-4-sec-12.8-solns

hw-4-sec-12.8-solns - SECTION 12.8 POWER SERIES 991...

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Unformatted text preview: SECTION 12.8 POWER SERIES 991 . f“ /”‘ n n+1 Vthh is ’ ' 3 a : £— then lim an“ 2 lim :r -L—— ‘ l‘ ‘lml — U J77], n—voo an n—voo m \$n ”H00 n +177’LI20 1+1/7'L — [1:] 2 By the Ratio Test the series Z in converges when 1961 < 1, so the radius of convergence R: 1. Now we ’11 check the n n: 1 ) . _ 1 endpomts, that is, 3:: :1. When x— ~ 1, the series 2 — diverges because it is a 11— series with 19— ~ % g 1 When 90— — —1, n=1 Tl . 0° (—1)" . . . the series 2 \/_ converges by the Alternating Series Test. Thus, the interval of convergence is I = [—1, 1). 1y n21 Tl ‘1 n n . 11 . n+1 1 4.1fan : Qiﬁhen 11m a +1 : 11m x . 71+ — lim —L_ —|.\$| n+1 n—>oo an n—>oo n+2 HI" TL—>001+1/(n+1) . . 0° —1)"z” . . By the Ratio Test, the series 2 TH— converges when |m| < 1, so R = 1. When ac : —1, the series diverges because it n: is the harmonic series; when a: = 1, it is the alternating harmonic series, which converges by the Alternating Series Test. Thus, I : (—1,1]. _ 71.71 71 @fan 4( )3 30 ,then n n 1 n n+1 3 _1 3 3 lim (1 +1 2 lim i;— = lim (ﬂ = lim n |a:] =13-|m|: |x|.Bythe 77,—»00 an n—voo (TL + 1)3 (—1)n‘1;12” n—wo (n +1)3 71—00 77, +1 00 (_1)n‘lmn Ratio Test, the series 2 n3 converges when [m] < 1, so the radius of convergence R = 1. Now we’ll check the 71:1 00 _1): 1 endpoints, that is, a: = :1:1. When x = 1, the series ”Sf converges by the Alternating Series Test. When m = —1, - °° (—1) “1(4)" °° 1 the series 2 —3*— : — 3 converges because it is a constant multiple of a convergent p— series [10- W 3 > 1]. TL n— 1 TL — Thus, the interval of convergence is I = [—1, 1]. n+1 n V . 1 6. an : x/Ex", so we need lim (1 +1 71 1931+1 — 11m 1 + — |a:| = |as| < 1 for convergence (by the n—wo an nh—nso 7—1931” naoo TL Ratio Test), so R = 1. When a: = :: 1, lim 1an|= lirn \/7_1= 00, so the series diverges by the Test for Divergence. n—boo n—~><X> Thus, I = (~1,1). x" 7. If an 2 — then lim ’ 77,! 11—100 =1m|~0=0< lforallrealm. So, by the Ratio Test, R = 00 and I = (—00, oo). were the Root Test is easier. Ifan = 11"“then lirn \"/|an|— _ nlim n |m]— oo ifm 7a— 0, so R — 0 and I —— {0}. nwoo 712:5" 9. Ifan = (—1)" 2n ,then 2 . an+1 ~ . (n+1)2ar:"+1 2” , q . 20011-1)2 _ [ml 1 |x| 2 .1132, a, 3131;, —-—2.+1 ‘nzxn 21:12. 2.2 2132. 7 1+; -—2 <1) =21“ Bythe 1’ /’~, \2 ,N 1’7. 11? 992 CHAPTER 12 INFINITE SEQUENCES AND SERIES 2 n Ratio Test, the series 2 (—1)" n2: converges when — 2Incl < 1 4:} IxI < 2, so the radius of convergence is R = 2. 0° 2 :22 0° . . . When :1: 2 :2, both series 2 (51)" 712‘): “Z (4:1)"712 dlverge by the Test for D1vergence Since lim I(q:1)" n2I — 00. Thus, the interval ofconvergence is I : (—2, 2). 10. Ifan = 10 39: ,then n , an+1 , 10"+1 93"“ n3 . 1017713 10ImI 10le 1 z 1 . : 1 ~ 1 z 2 .520 a. ”in; (n + Us mm "220 1” +113 .320 (1+ W13 13 10m 0711: 00 By the Ratio Test, the series 7111: "converges when 10 IxI < 1 ¢> - - ~ 1 ILCI <—1—0, so the radius ofconvergence 18 R — 10- When a:— ~ ~ﬁ, the series converges by the Alternating Series Test; when :12: To , the series converges because it is a [7— series ~ . _ 1 1 W1th p = 3 > 1. Thus, the interval of convergence 1s I — I—ﬁ, ﬁl- ~2 n n n . 27L+1 n+1 4 11. an 2 (4, so lim (2 +1 2 11m IQCI f" = lim 2I\$I4 * 2—ImI, so by the Ratio Test, the v; M. an M, w +1 2711 I W. . . 0° 1 series converges when 2 IxI < 1 1:} ImI < 1, so R : %. When ac 2 —%, we get the d1vergentp—ser1es 2 71:1 71 F II NH I/\ 1 . When :10 : %, we get the series 2 (‘ n: , which converges by the Alternating Series Test. é/ﬁ , . :L‘” . @0471 : 5n n5 ’ SO "1:120 n+1 n 5 5 an+1 _ , m 5 n ¥ IxI n _ IarI . . an — 1m;o m - m" _ 711320 F n + 1 _ ? By the Ratlo Test, the series °° as" [art _ _ _ . oo (—11" . Z 5 5 converges when E < 1 (it ImI < 5, so R ~ 5. When 1 _ 5, we get the serles E n5 , Wthh 71:0 71:1 converges by the Alternating Series Test. When a: = 5, we get the convergent p—series 2 71—15 [1) = 5 > 1]. n=1 Thus, I = [—5, 5]. n n+1 7L . an+1 . a: 4 lnn ILL‘I , lnn IazI . f n 2 —1 n ’t l : l . : — 1 z — - 13 I a I ) lnn he“ 71330 an "3110 4n+11n(n + 1) :17" 4 4 1 n—wo ln(n -I— 1) [byl’ Hospital’ s Rule]: %. By the Ratio Test, the senes converges when % < 1 {I} IxI<4,soR=4. When _ * > _ __ 4, ”E 2-( 1) n1 :E2\n1: n E21.Slncelnn<nforn 2,1 > nand "(€021 181116 divergent harmonic series (without the n — 1 term) n=2 1:271 2n+2 2 . n+1 ILL’I (2n)! . IwI n 1 - — ___ § : 1 _ _ _ t 14-‘1 ‘(1)n(2n)1950n1£§o an —n1111.10(2n+2)! M2” T131010 (271+ 1)(2nn+2)__0 < 1. Thus, by he Ratio Test, the series converges for all real .1? and we have R : 00 and I = (—00, oo). SECTION12.8 POWER SERIES 3 993 (\$22)” . an+1 ' (I‘2)n+l ”2+1 . ”2+1 15.Ifn2—~—,th l _l ————‘——2 ~21 “—2 —, (1 n2 + 1 en nLngo a,” 71320 (n + 1)2 +1 (1' h 2)” lm lug/1010 (TL +1)2 +1 ].T 2| By the . . °° (m2 2)" Ratio Test, the senes Z —TLZ—+—1 converges when lac ~ 2] <1 [R 2 1] <2 ~1 < an ~ 2 < 1 <2 1 < a: < 3. When n20 . 0° ,, 1 . . . . 0° 1 x 2 1, the series TEES—1) n2 + 1 converges by the Alternating Ser1es Test, when a: 2 3, the serles "go n2 + 1 converges by . . . °° 1 . . comparison With the p-ser1es 2 E [p 2 2 > 1]. Thus, the 1nterval of convergence is I 2 [1,3]. 7121 (3: ~ 3)" an+1 , (m « 3)"+1 2n +1 . 2n + 1 ’16fn2~" ,hl 21 « _- 2—. / a ( ) 271+ 1 en 7.12120 an ”in; 2n+3 (ac A 3)” II \$711520 2n+3 Im 3| By the . . °° — 3 n Ratio Test, the senes Z (—1)” (élﬁ converges when ]\$ — 3] < 1 [R 2 1] <2 —1 < :15 — 3 <1 <2 2 < 1‘ < 4. n20 . loo 1 . . . . . . . When \$ 2 2, the series 2 2n 1 dlverges by limlt comparlson w1th the harmonlc series (or by the Integral Test); when ”:0 “— . 0° 1 . . . . a: 2 4, the series 2 (—1)” 2n 1 converges by the Alternatmg Series Test. Thus, the interval of convergence IS I 2 (2, 4]. n=0 __ n 4 n n n+1 4 n+1 17 fan 2 3 (”5+ ) ,then 11111 a +1 — lim 3 (36+) . ‘5 _3|a:+4] lim ‘5 23|m+4|. x/E "20° an "2‘” n +1 3"(\$ + 4)" "200 n + l BhR'T h 'm3n(m+4)n h3 4 1 4 1 —1 yt e atlo est,t eseries 7&1 T convergesw en |\$ + | < 42 ]\$+ | < 5 [R 2 3] <2 . °° 1 . . _% < a: + 4 < % <2 —l3§ < a: < —%. When :3 2 —i3‘°3, the ser1es 225—1)" E converges by the Alternatlng Series Test; when a: 2 —%, the series 21 AT; diverges [p 2 % g 1]. Thus, the interval of convergence is I 2 [—13% WE). _ n n . an“ _ . (n+1)(ar:+1)"+1 4” ]m+1l . n+1_]\$+1] 18. Ifan — 21: (m + 1) ,then "lingo an "1520 4n+1 . ”(1” +1)” _ 4 ”lingo n 2 4 . By the Ratio Test, the series 2 41" (ac + 1)" converges when ‘11 :— 1! < 1 <2 |m + 1] < 4 [R = 4] <2 7121 —4 < a: + 1 < 4 <2 —5 < m < 3. When .1‘ 2 —5 or 3, both series 2 (:1)" n diverge by the Test for Divergence since n21 lim |(:{:1)" n] 2 00. Thus, the interval ofconvergence is I 2 (—5, 3). 1:“ an : M, then lim ‘"/ Ian] : lim |x A 2t 2 0, so the series converges for all m (by the Root Test). 77," n—>oo n—>oo 'n, R2 ooandI 2 (—oo,oo). . an+1 . (3a: — 2)"+1 n3" . |3a3 — 2] 1 |3m — 2| 2 , . = 1 — . —— = 1 . z = 2 — , b th R 2° ”1320 an ninolo (n + 1) 371+1 (32: — 2)” mimic 3 1 + l/n 3 I“ 3] 5° y e at“) ~ 2 1 5 1 - ~ 0° (“U11 Test, the serles converges when ]m — 5] < 1 (2 —5 < ac < 3. R 2 1. When a; 2 —3, the ser1es15 Z , the 7121 TL convergent alternating harmonic series. When an 2 g, the series becomes the divergent harmonic series. Thus, I 2 [—%, g) <1 I 2 1 \(r 994 CHAPTER 12 INFINITE SEQUENCES AND SERIES n TL 21. an : 1170: ~ a) ,where b > 0. an+1 an lim TL—>Cx) "goo bnI—l __ ”+1 71 _hm(n+1)lrr a; , b < . . — a By the Ratlo Test, the series converges when Ix I <1 © Ix—a|<b [sosz] e» —b<a:—a<b 4: (1 ~ I) < x < a + b. When [:2 — aI— a 1), lim IanI— — lim n : 00, so the series diverges. Thus, I : (a — b, a + b). 'J/ \ 71(3: ~ 4)" . an“ . (n+1) I\$~4In+1 n3+1 , 1 n3+1 l : l K.‘ : 1 1 — -4 : _ . ningo an 11520 (n + 1)3 +1 72 Im — 4|" 71320 + n3 —I— 3712 + 3n + 2 Iac I I3: 4' By the Ratio Test, the series converges when [:27 — 4I < 1 [so R z 1] <¢ —1 < m — 4 < 1 ® 3 < a: < 5. When 00 In: — 4I : 1, :1 IanI = Z 1137:— 1 , which converges by comparison with the convergent p-series Z p = 2 > 1]. Thus, I = [3, 5] . n I 2 _ n+1 23.1fm, :n! (Zr—l)”, then lim a +1 : lim (n+1) ( a: 1) —lim~ (n+1)I2x—1I—>ooasn—>oo n-—>oo an nﬂoo TZI(21‘ — 1)" n—>oo for all x 75 5 Since the series diverges for all x aé %, R : 0 and I = {%}. 24 _ 7123:” 7122:" 711:" so “a” 2-4-6 ----- (2n) * 2w “Mn—w n+1 n n . 2 — 1 I , 1 . . lim (1 +1 2 11m (n + 1) I33I -—£L‘): 11m 71 + m = 0. Thus, by the Ratio Test, the series converges for n—roo an n—>oo 2n+lnI7L71I\$I nﬁoo 7L2 2 all real 9: and we have R : 00 and I = (—00, 00). n+1 2 n . . 1 . , 25_ lim (1 +1 : Tim [M— - ‘n—‘ﬁI : 11m I41, + I2 2 I417 + 1I, so by the Ratio Test, the series converges when I4\$ + 1] <1 4:) ~1 < 4m+1< 1 <:> —2<4m<0 <=> —%<x<0,s0R=%.Whenzr=—~;, _ TL the series becomes HZ ( ”1) =1 ,which converges by the Alternating Series Test. When as = 0, the series becomes 2 —12—, M: in? a convergent p-series [p = 2 > 1]. I = I~%, OI. 232" a +1 332”” n (1n n)2 n (In n)2 . n 2 “—— t 1 n : 1 . ‘ 2 N z 2' 26 I” n (1nn)2’ he“ "in; an 1.31;, (n + 1)[ln(n + 1)]2 17271:” I313; (n +1)[ln(n + 1)]2 ”3 00 (132” By the Ratio Test, the series 2: m converges when m2 < 1 4:} ImI < 1, so R = 1. When a: = :I:1, \$271 2‘ 1’ the 71:2 TL 77' °° 1 series 2 W converges by the Integral Test (see Exercise 12.3.22). Thus, the interval of convergence is I = I ”:2 n nn 4,1]. / ...
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