Solucao impares Calculo 6ed CAP.11

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(1)F. TX.10. 11. INFINITE SEQUENCES AND SERIES. 11.1 Sequences 1. (a) A sequence is an ordered list of numbers. It can also be dened as a function whose domain is the set of positive integers.. (b) The terms dq approach 8 as q becomes large. In fact, we can make dq as close to 8 as we like by taking q sufciently large. (c) The terms dq become large as q becomes large. In fact, we can make dq as large as we like by taking q sufciently large. q. 3. dq = 1 (0=2) , so the sequence is {0=8> 0=96> 0=992> 0=9984> 0=99968> = = =}. 5. dq =. 3 (1)q , so the sequence is q!. 7. d1 = 3, dq+1 = 2dq 1.. . 3 1 1 1 3 3 3 3 3 > > > > > = = = = 3> > > > > = = = . 1 2 6 24 120 2 2 8 40. Each term is dened in terms of the preceding term.. d2 = 2d1 1 = 2(3) 1 = 5. d3 = 2d2 1 = 2(5) 1 = 9. d4 = 2d3 1 = 2(9) 1 = 17. d5 = 2d4 1 = 2(17) 1 = 33. The sequence is {3> 5> 9> 17> 33> = = =}. 9.. 1 1 1 1

(2) 1> 3 > 5 > 7 > 9 > = = = . The denominator of the nth term is the nth positive odd integer, so dq =. 11. {2> 7> 12> 17> = = =}. 13.. 1 . 2q 1. Each term is larger than the preceding one by 5, so dq = d1 + g(q 1) = 2 + 5(q 1) = 5q 3..

(3) q1. 8 > = = = . Each term is 23 times the preceding one, so dq = 23 . 1> 23 > 49 > 27. 15. The rst six terms of dq =. lim. q. 1 2 3 4 5 6 1 q are , , , , , . It appears that the sequence is approaching . 2q + 1 3 5 7 9 11 13 2. q 1 1 = lim = 2q + 1 q 2 + 1@q 2. 17. dq = 1 (0=2)q , so lim dq = 1 0 = 1 by (9). q. 19. dq =. Converges. 3 + 5q2 (3 + 5q2 )@q2 5 + 3@q2 5+0 , so dq = 5 as q . Converges = = q + q2 (q + q2 )@q2 1 + 1@q 1+0. 21. Because the natural exponential function is continuous at 0, Theorem 7 enables us to write. lim dq = lim h1@q = hlimq$4 (1@q) = h0 = 1= Converges. q. q. (2q)@q 2 2 2q , then lim eq = lim = lim = = . Since tan is continuous at q q (1 + 8q)@q q 1@q + 8 1 + 8q 8 4 2q 2q Theorem 7, lim tan = tan lim = tan = 1. Converges q q 1 + 8q 1 + 8q 4. 23. If eq =. 4,. by. 455.

(4) F. 456. ¤. 25. dq =. TX.10. INFINITE SEQUENCES AND SERIES. CHAPTER 11. (1)q1 1 (1)q1 q 1 = , so 0 |dq | = 0 as q , so dq 0 by the Squeeze Theorem and 2 q +1 q + 1@q q + 1@q q. Theorem 6. Converges 27. dq = cos( q@2).. This sequence diverges since the terms don’t approach any particular real number as q .. The terms take on values between 1 and 1= 29. dq =. (2q 1)! 1 (2q 1)! = = 0 as q . Converges (2q + 1)! (2q + 1)(2q)(2q 1)! (2q + 1)(2q). 31. dq =. hq + hq hq 1 + h2q · = 0 as q because 1 + h2q 1 and hq hq . Converges h2q 1 hq hq hq. {2 H 2{ H 2 q2 . Since lim { = lim { = lim { = 0, it follows from Theorem 3 that lim dq = 0. Converges { h { h { h q hq 2 cos2 q 1 cos q 1 2 35. 0 [since 0 cos q 1], so since lim = 0, converges to 0 by the Squeeze Theorem. q 2q 2q 2q 2q 33. dq = q2 hq =. 37. dq = q sin(1@q) =. sin(1@q) sin(1@{) sin w . Since lim = lim [where w = 1@{] = 1, it follows from Theorem 3 { 1@q 1@{ w w0+. that {dq } converges to 1. 2 , so ln | = { ln 1 + { 1 2 2 1 + 2@{ { ln(1 + 2@{) H 2 =2 lim ln | = lim = lim = lim 2 { { { { 1 + 2@{ 1@{ 1@{ { q 2 2 = lim hln | = h2 , so by Theorem 3, lim 1 + = h2 . Convergent lim 1 + { { q { q. 39. | =. { 2 1+ {. 41. dq = ln(2q2 + 1) ln(q2 + 1) = ln. . 2q2 + 1 q2 + 1. . = ln. 2 + 1@q2 1 + 1@q2. ln 2 as q . Convergent. 43. {0> 1> 0> 0> 1> 0> 0> 0> 1> = = =} diverges since the sequence takes on only two values, 0 and 1, and never stays arbitrarily close to. either one (or any other value) for q sufciently large. 45. dq = 47.. (q 1) q 1 q q! 1 2 3 · · = · · · ··· · 2q 2 2 2 2 2 2 2. [for q A 1] =. q as q , so {dq } diverges. 4. From the graph, it appears that the sequence converges to 1. {(2@h)q } converges to 0 by (9), and hence {1 + (2@h)q } converges to 1 + 0 = 1..

(5) F. TX.10. SECTION 11.1. ¤. SEQUENCES. 457. From the graph, it appears that the sequence converges to 12 .. 49.. As q , v u 3 + 2q2 3@q2 + 2 dq = = 2 8q + q 8 + 1@q. u . 0+2 = 8+0. u. 1 1 = , 4 2. so lim dq = 12 . q. From the graph, it appears that the sequence {dq } =. 51.. q2 cos q 1 + q2. is. divergent, since it oscillates between 1 and 1 (approximately). To prove this, suppose that {dq } converges to O. If eq = {eq } converges to 1, and lim. q. lim. q. q2 , then 1 + q2. dq dq O = O. But = = cos q, so eq 1 eq. dq does not exist. This contradiction shows that {dq } diverges. eq. From the graph, it appears that the sequence approaches 0.. 53.. 0 ? dq =. 3 5 2q 1 1 · 3 · 5 · · · · · (2q 1) 1 · · · ··· · = (2q)q 2q 2q 2q 2q. 1 1 · (1) · (1) · · · · · (1) = 0 as q 2q 2q 1 · 3 · 5 · · · · · (2q 1) So by the Squeeze Theorem, converges to 0. (2q)q . 55. (a) dq = 1000(1=06)q. d1 = 1060, d2 = 1123=60, d3 = 1191=02, d4 = 1262=48, and d5 = 1338=23.. (b) lim dq = 1000 lim (1=06)q , so the sequence diverges by (9) with u = 1=06 A 1. q. q. 57. If |u| 1, then {uq } diverges by (9), so {quq } diverges also, since |quq | = q |uq | |uq |. If |u| ? 1 then. lim {u{ = lim. {. {. { u{. H. 1 u{ = 0, so lim quq = 0, and hence {quq } converges = lim { ( ln u) u { { ln u q. = lim. whenever |u| ? 1. 59. Since {dq } is a decreasing sequence, dq A dq+1 for all q 1. Because all of its terms lie between 5 and 8, {dq } is a. bounded sequence. By the Monotonic Sequence Theorem, {dq } is convergent; that is, {dq } has a limit O. O must be less than 8 since {dq } is decreasing, so 5 O ? 8. 61. dq =. 1 1 1 1 is decreasing since dq+1 = = ? = dq for each q 1. The sequence is 2q + 3 2(q + 1) + 3 2q + 5 2q + 3. bounded since 0 ? dq . 1 5. for all q 1. Note that d1 = 15 ..

(6) F. 458. ¤. TX.10. INFINITE SEQUENCES AND SERIES. CHAPTER 11. 63. The terms of dq = q(1)q alternate in sign, so the sequence is not monotonic. The rst ve terms are 1, 2, 3, 4, and 5.. Since lim |dq | = lim q = , the sequence is not bounded. q. q. 2 { + 1 (1) {(2{) { q 1 {2 0 denes a decreasing sequence since for i ({) = 2 , i ({) = 65. dq = 2 = 0 2 2 q +1 { +1 ({ + 1) ({2 + 1)2 for { 1. The sequence is bounded since 0 ? dq 67. For. 1 2. for all q 1.. t s s q q q 2, 2 2, 2 2 2, = = = , d1 = 21@2 , d2 = 23@4 , d3 = 27@8 , = = =, so dq = 2(2 1)@2 = 21(1@2 ) . q). lim dq = lim 21(1@2. q. q. = 21 = 2.. Alternate solution: Let O = lim dq . (We could show the limit exists by showing that {dq } is bounded and increasing.) q. Then O must satisfy O = 2 · O O2 = 2O O(O 2) = 0. O 6= 0 since the sequence increases, so O = 2. 69. d1 = 1, dq+1 = 3 . 1 . We show by induction that {dq } is increasing and bounded above by 3. Let Sq be the proposition dq. that dq+1 A dq and 0 ? dq ? 3. Clearly S1 is true. Assume that Sq is true. Then dq+1 A dq . 1 1 1 1 A . Now dq+2 = 3 A3 = dq+1 dq+1 dq dq+1 dq. . 1 1 ? dq+1 dq. . Sq+1 . This proves that {dq } is increasing and bounded. above by 3, so 1 = d1 ? dq ? 3, that is, {dq } is bounded, and hence convergent by the Monotonic Sequence Theorem. If O = lim dq , then lim dq+1 = O also, so O must satisfy O = 3 1@O O2 3O + 1 = 0 O = q. q. But O A 1, so O =. 3± 5 . 2. 3+ 5 . 2. 71. (a) Let dq be the number of rabbit pairs in the nth month. Clearly d1 = 1 = d2 . In the nth month, each pair that is. 2 or more months old (that is, dq2 pairs) will produce a new pair to add to the dq1 pairs already present. Thus, dq = dq1 + dq2 , so that {dq } = {iq }, the Fibonacci sequence. (b) dq =. iq+1 iq. dq1 =. iq iq1 + iq2 iq2 1 1 = =1+ =1+ =1+ . If O = lim dq , q iq1 iq1 iq1 iq1 /iq2 dq2. then O = lim dq1 and O = lim dq2 , so O must satisfy O = 1 + q. q. 1 O. O2 O 1 = 0 O =. [since O must be positive]. 73. (a). From the graph, it appears that the sequence q5 = 0. q q!. converges to 0, that is, lim. q5 q!. . 1+ 5 2.

(7) F. TX.10. SECTION 11.1. SEQUENCES. ¤. 459. (b). From the rst graph, it seems that the smallest possible value of Q corresponding to % = 0=1 is 9, since q5 @q! ? 0=1 whenever q 10, but 95 @9! A 0=1. From the second graph, it seems that for % = 0=001, the smallest possible value for Q is 11 since q5 @q! ? 0=001 whenever q 12. 75. Theorem 6: If lim |dq | = 0 then lim |dq | = 0, and since |dq | dq |dq |, we have that lim dq = 0 by the q. q. q. Squeeze Theorem. 77. To Prove: If lim dq = 0 and {eq } is bounded, then lim (dq eq ) = 0. q. q. Proof: Since {eq } is bounded, there is a positive number P such that |eq | P and hence, |dq | |eq | |dq | P for all q 1. Let % A 0 be given. Since lim dq = 0, there is an integer Q such that |dq 0| ? q. |dq eq 0| = |dq eq | = |dq | |eq | |dq | P = |dq 0| P ?. % if q A Q. Then P. % · P = % for all q A Q. Since % was arbitrary, P. lim (dq eq ) = 0.. q. 79. (a) First we show that d A d1 A e1 A e.. d1 e1 =. d+e 2. . 2 de = 12 d 2 de + e = 12 d e A 0 [since d A e] d1 A e1 . Also. d d1 = d 12 (d + e) = 12 (d e) A 0 and e e1 = e . de = e e d ? 0, so d A d1 A e1 A e. In the same. way we can show that d1 A d2 A e2 A e1 and so the given assertion is true for q = 1. Suppose it is true for q = n, that is, dn A dn+1 A en+1 A en . Then dn+2 en+2 = 12 (dn+1 + en+1 ) . 2 s s s dn+1 en+1 = 12 dn+1 2 dn+1 en+1 + en+1 = 12 dn+1 en+1 A 0,. dn+1 dn+2 = dn+1 12 (dn+1 + en+1 ) = 12 (dn+1 en+1 ) A 0, and en+1 en+2 = en+1 . s s s dn+1 en+1 = en+1 en+1 dn+1 ? 0. . dn+1 A dn+2 A en+2 A en+1 ,. so the assertion is true for q = n + 1. Thus, it is true for all q by mathematical induction. (b) From part (a) we have d A dq A dq+1 A eq+1 A eq A e, which shows that both sequences, {dq } and {eq }, are monotonic and bounded. So they are both convergent by the Monotonic Sequence Theorem. (c) Let lim dq = and lim eq = . Then lim dq+1 = lim q. 2 = + . q. = .. q. q. dq + eq 2. =. + 2. .

(8) F. 460. ¤. CHAPTER 11. INFINITE SEQUENCES AND SERIES. 81. (a) Suppose {sq } converges to s. Then sq+1 =. s2 + ds = es. esq d + sq. TX.10 . lim sq+1 =. q. e lim sq q. d + lim sq. . s=. q. es d+s. . . s(s + d e) = 0 s = 0 or s = e d. e sq d e esq sq ? (b) sq+1 = sq since 1 + A 1. = sq d + sq d d 1+ d 2 3 q e e e e e e (c) By part (b), s1 ? s0 , s3 ? s0 , etc. In general, sq ? s0 , s0 , s2 ? s1 ? s2 ? d d d d d d q e e · s0 = 0 since e ? d. By result 9> lim uq = 0 if 1 ? u ? 1. Here u = (0> 1) . so lim sq lim q q q d d (d) Let d ? e. We rst show, by induction, that if s0 ? e d, then sq ? e d and sq+1 A sq . For q = 0, we have s1 s0 =. es0 s0 (e d s0 ) s0 = A 0 since s0 ? e d. So s1 A s0 . d + s0 d + s0. Now we suppose the assertion is true for q = n, that is, sn ? e d and sn+1 A sn . Then e d sn+1 = e d . esn d(e d) + esn dsn esn d(e d sn ) = = A 0 because sn ? e d. So d + sn d + sn d + sn. sn+1 ? e d. And sn+2 sn+1 =. esn+1 sn+1 (e d sn+1 ) sn+1 = A 0 since sn+1 ? e d. Therefore, d + sn+1 d + sn+1. sn+2 A sn+1 . Thus, the assertion is true for q = n + 1. It is therefore true for all q by mathematical induction. A similar proof by induction shows that if s0 A e d, then sq A e d and {sq } is decreasing. In either case the sequence {sq } is bounded and monotonic, so it is convergent by the Monotonic Sequence Theorem. It then follows from part (a) that lim sq = e d. q. 11.2 Series 1. (a) A sequence is an ordered list of numbers whereas 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. 3.. q. vq. 1. 2=40000. 2. 1=92000. 3. 2=01600. 4. 1=99680. 5. 2=00064. 6. 1=99987. 7. 2=00003. 8. 1=99999. From the graph and the table, it seems that the series converges to 2. In fact, it is a geometric. 9. 2=00000. 10. 2=00000. series with d = 2=4 and u = 15 , so its sum is. S q=1. 12 2=4 2=4 = = 2= = (5)q 1=2 1 15. Note that the dot corresponding to q = 1 is part of both {dq } and {vq }. TI-86 Note: To graph {dq } and {vq }, set your calculator to Param mode and DrawDot mode. (DrawDot is under.

(9) F. TX.10. SECTION 11.2. SERIES. ¤. GRAPH, MORE, FORMT (F3).) Now under E(t)= make the assignments: xt1=t, yt1=12/(-5)ˆt, xt2=t, yt2=sum seq(yt1,t,1,t,1). (sum and seq are under LIST, OPS (F5), MORE.) Under WIND use 1,10,1,0,10,1,-3,1,1 to obtain a graph similar to the one above. Then use TRACE (F4) to see the values. 5.. q. vq. 1. 1=55741. 2. 0=62763. 3. 0=77018. 4. 0=38764. 5. 2=99287. 6. 3=28388. 7. 2=41243. 8. 9=21214. 9. 9=66446. 10. 9=01610. The series. S q=1. tan q diverges, since its terms do not approach 0=. 7.. q. vq. 1. 0=29289. 2. 0=42265. 3. 0=50000. 4. 0=55279. 5. 0=59175. 6. 0=62204. 7. 0=64645. 8. 0=66667. 9. 0=68377. 10. 0=69849. 9. (a) lim dq = lim q. q. (b) Since lim dq = q. 11. 3 + 2 +. 4 3. +. converges to 13. 3 4 +. 16 3. 8 9. 2q 2 = , so the sequence {dq } is convergent by (11.1.1). 3q + 1 3 2 3. 6= 0, the series. S q=1. dq is divergent by the Test for Divergence.. + · · · is a geometric series with rst term d = 3 and common ratio u = 23 . Since |u| =. d 1u. . From the graph and the table, it seems that the series converges. n S 1 1 1 1 1 1 1 1 = + + ··· + q q+1 n+1 1 2 2 3 n q=1 1 =1 , n+1 S 1 1 1 = lim 1 = 1. so n q q+1 n+1 q=1. 64 9. =. 3 12@3. =. 3 1@3. = 9.. + · · · is a geometric series with ratio u = 43 . Since |u| =. 4 3. A 1, the series diverges.. 2 3. ? 1, the series. 461.

(10) F. 462. 15.. ¤ S q=1. CHAPTER 11. TX.10. INFINITE SEQUENCES AND SERIES. 6(0=9)q1 is a geometric series with rst term d = 6 and ratio u = 0=9. Since |u| = 0=9 ? 1, the series converges to. d 6 6 = = = 60. 1u 1 0=9 0=1 17.. q1 (3)q1 S 1 S 3 = . The latter series is geometric with d = 1 and ratio u = 34 . Since |u| = 4q 4 q=1 4 q=1 converges to. 19.. S q=0. 21.. 3 4. ? 1, it. 1 = 47 . Thus, the given series converges to 14 47 = 17 . 1 (3@4). q q 1 S = is a geometric series with ratio u = . Since |u| A 1, the series diverges. 3q+1 3 q=0 3 3. 1 1 S 1 S = diverges since each of its partial sums is 12 times the corresponding partial sum of the harmonic series 2 q=1 q q=1 2q 1 1 S S 1 S , which diverges. If were to converge, then would also have to converge by Theorem 8(i)= q=1 q q=1 2q q=1 q. In general, constant multiples of divergent series are divergent. 23.. S n=2. n2 n2 diverges by the Test for Divergence since lim dn = lim 2 = 1 6= 0. n n n 1 1. n2. 25. Converges.. q q 1 + 2q S S S 2q 2 1 1 = + + = q 3q 3q 3 3 q=1 q=1 3 q=1 =. 27.. lim dq = lim. q. 31.. q. q 2 = lim 21@q = 20 = 1 6= 0. q. 2 q +1 ln diverges by the Test for Divergence since 2q2 + 1 q=1 2 q2 + 1 q +1 = ln lim = ln 12 6= 0. lim dq = lim ln q q q 2q2 + 1 2q2 + 1 S. S q=1. 33.. 2@3 1 5 1@3 + = +2= 1 1@3 1 2@3 2 2. S q 2 = 2 + 2 + 3 2 + 4 2 + · · · diverges by the Test for Divergence since. q=1. 29.. [sum of two convergent geometric series]. arctan q diverges by the Test for Divergence since lim dq = lim arctan q =. 1 S S = q h q=1 q=1. q. q. 2. 6= 0.. q 1 1 1 1 is a geometric series with rst term d = and ratio u = . Since |u| = ? 1, the series converges h h h h. S 1 1@h h 1 1@h = · = . By Example 6, = 1. Thus, by Theorem 8(ii), 1 1@h 1 1@h h h1 q=1 q(q + 1) 1 S S S 1 1 1 1 h1 h 1 = = +1 = + = . + + q q h q(q + 1) h1 h1 h1 h1 q=1 q=1 h q=1 q(q + 1). to.

(11) F. TX.10. SECTION 11.2. SERIES. ¤. 463. S. 2 are 2 1 q q=2 q q S S 2 1 1 = vq = l1 l+1 l=2 (l 1)(l + 1) l=2 1 1 1 1 1 1 1 1 1 = 1 + + + ··· + + 3 2 4 3 5 q3 q1 q2 q. 35. Using partial fractions, the partial sums of the series. 1 1 1 . 2 q1 q S 1 2 1 1 3 = lim vq = lim 1 + = . Thus, 2 1 q q q 2 q 1 q 2 q=2. This sum is a telescoping series and vq = 1 +. q q S S 3 3 1 1 , vq = = [using partial fractions]. The latter sum is l+3 q=1 q(q + 3) l=1 l(l + 3) l=1 l 1 1 1 1 q1 + q 12 q + q+2 1 14 + 12 15 + 13 16 + 14 17 + · · · + q3 + q1 + q1 1 S. 37. For the series. =1+ Thus,. 3 = lim vq = lim 1 + q q q=1 q(q + 3) S. 39. For the series. 1 2. +. 1 3. . 1 q +1. . 1 2. 1 q+2. + . 1 3. . 1 q+3. 1 q +1. . . 1 q+2. =1+. 1 2. . +. 1 3. 1 q+3. =. 1 q+3. . [telescoping series]. 11 . 6. Converges. S h1@q h1@(q+1) ,. q=1. q S h1@l h1@(l+1) = (h1 h1@2 ) + (h1@2 h1@3 ) + · · · + h1@q h1@(q+1) = h h1@(q+1) vq = l=1. S Thus, h1@q h1@(q+1) = lim vq = lim h h1@(q+1) = h h0 = h 1. Converges q. q=1. 41. 0=2 =. q. 2@10 2 1 d 2 2 2 + 2 + · · · is a geometric series with d = and u = . It converges to = = . 10 10 10 10 1u 1 1@10 9. 43. 3=417 = 3 +. 417 417 417 417 417 1 + 6 + · · · . Now 3 + 6 + · · · is a geometric series with d = 3 and u = 3 . 103 10 10 10 10 10. It converges to. 417@103 3414 1138 d 417@103 417 417 = . Thus, 3=417 = 3 + = = . = = 3 1u 1 1@10 999@103 999 999 999 333. 45. 1=5342 = 1=53 +. It converges to. 42 42 42 42 42 1 + 6 + · · · . Now 4 + 6 + · · · is a geometric series with d = 4 and u = 2 . 104 10 10 10 10 10. 42@104 42@104 42 d = . = = 2 1u 1 1@10 99@102 9900. Thus, 1=5342 = 1=53 + 47.. [telescoping series]. 153 42 15,147 42 15,189 5063 42 = + = + = or . 9900 100 9900 9900 9900 9900 3300. {q { q S S { = is a geometric series with u = , so the series converges |u| ? 1 q 3 q=1 3 q=1 3. that is, 3 ? { ? 3. In that case, the sum of the series is. |{| ? 1 |{| ? 3; 3. {@3 {@3 3 { d = = · = . 1u 1 {@3 1 {@3 3 3{.

(12) F. 464. 49.. ¤ S q=0. 4q {q =. S q=0. (4{)q is a geometric series with u = 4{, so the series converges |u| ? 1 4 |{| ? 1 . |{| ? 14 . In that case, the sum of the series is 51.. TX.10. INFINITE SEQUENCES AND SERIES. CHAPTER 11. 1 . 1 4{. cosq { S cos { , so it converges is a geometric series with rst term 1 and ratio u = q 2 2 q=0. for all {. Thus, the series converges for all real values of { and the sum of the series is. |u| ? 1. But |u| =. |cos {| 1 2 2. 2 1 = . 1 (cos {)@2 2 cos {. 53. After dening i, We use convert(f,parfrac); in Maple, Apart in Mathematica, or Expand Rational and. 3q2 + 3q + 1 1 1 = 3 . So the nth partial sum is (q2 + q)3 q (q + 1)3 q S 1 1 1 1 1 1 1 1 = 1 + + · · · + =1 vq = 3 (n + 1)3 23 23 33 q3 (q + 1)3 (q + 1)3 n=1 n. Simplify in Derive to nd that the general term is. The series converges to lim vq = 1. This can be conrmed by directly computing the sum using sum(f,1..infinity); q. (in Maple), Sum[f,{n,1,Infinity}] (in Mathematica), or Calculus Sum (from 1 to ) and Simplify (in Derive). 55. For q = 1, d1 = 0 since v1 = 0. For q A 1,. dq = vq vq1 = Also,. S q=1. dq = lim vq = lim q. q. (q 1) 1 (q 1)q (q + 1)(q 2) 2 q1 = = q+1 (q 1) + 1 (q + 1)q q(q + 1). 1 1@q = 1. 1 + 1@q. 57. (a) The rst step in the chain occurs when the local government spends G dollars. The people who receive it spend a fraction f. of those G dollars, that is, Gf dollars. Those who receive the Gf dollars spend a fraction f of it, that is, Gf2 dollars. Continuing in this way, we see that the total spending after q transactions is Vq = G + Gf + Gf2 + · · · + Gfq–1 = (b) lim Vq = lim q. q. G = v. G(1 fq ) by (3). 1f. G(1 fq ) G G = lim (1 fq ) = 1f 1 f q 1f. [since f + v = 1] = nG. k since 0 ? f ? 1. . l lim fq = 0. q. [since n = 1@v]. If f = 0=8, then v = 1 f = 0=2 and the multiplier is n = 1@v = 5. 59.. S. (1 + f)q is a geometric series with d = (1 + f)2 and u = (1 + f)1 , so the series converges when. q=2. (1 + f)1 ? 1 |1 + f| A 1 1 + f A 1 or 1 + f ? 1 f A 0 or f ? 2. We calculate the sum of the. series and set it equal to 2:. (1 + f)2 =2 1 (1 + f)1. 2f2 + 2f 1 = 0 f = So f =. . 31 . 2. 2 ± 4. 12. =. ± 31 . 2. . 1 1+f. 2. 1 =22 1 = 2(1 + f)2 2(1 + f) 1+f. However, the negative root is inadmissible because 2 ?. 31 2. ? 0..

(13) F. TX.10 1. 1. . 1. SECTION 11.2. 61. hvq = h1+ 2 + 3 +···+ q = h1 h1@2 h1@3 · · · h1@q A (1 + 1) 1 +. =. 1 2. 1 + 13 · · · 1 + q1. SERIES. ¤. 465. [h{ A 1 + {]. 234 q+1 ··· =q+1 123 q. Thus, hvq A q + 1 and lim hvq = . Since {vq } is increasing, lim vq = , implying that the harmonic series is q. q. divergent. 63. Let gq be the diameter of Fq . We draw lines from the centers of the Fl to. the center of G (or F), and using the Pythagorean Theorem, we can write 2 2 . 12 + 1 12 g1 = 1 + 12 g1 2 2 1 = 1 + 12 g1 1 12 g1 = 2g1 [difference of squares] g1 = 12 . Similarly, 2 2 1 = 1 + 12 g2 1 g1 12 g2 = 2g2 + 2g1 g21 g1 g2 = (2 g1 )(g1 + g2 ) g2 = gq+1. 2 2 1 (1 g1 )2 [1 (g1 + g2 )]2 , and in general, g1 = , 1 = 1 + 12 g3 1 g1 g2 12 g3 g3 = 2 g1 2 g1 2 (g1 + g2 ) 2 S 1 q 1 1 l=1 gl S and = . If we actually calculate g2 and g3 from the formulas above, we nd that they are = 2 q g 6 2 ·3 l=1 l. 1 1 1 = respectively, so we suspect that in general, gq = . To prove this, we use induction: Assume that for all 12 3·4 q(q + 1) q S 1 1 q 1 1 = . Then = [telescoping sum]. Substituting this into our gl = 1 n(n + 1) n n+1 q+1 q+1 l=1 2 1 q 1 q+1 (q + 1)2 1 = formula for gq+1 , we get gq+1 = , and the induction is complete. = q+2 (q + 1)(q + 2) q 2 q+1 q+1 Sq Now, we observe that the partial sums l=1 gl of the diameters of the circles approach 1 as q ; that is,. n q, gn =. S q=1. dq =. S q=1. 1 = 1, which is what we wanted to prove. q(q + 1). 65. The series 1 1 + 1 1 + 1 1 + · · · diverges (geometric series with u = 1) so we cannot say that. 0 = 1 1 + 1 1 + 1 1 + ···. 67.. S. q=1. fdq = lim. q. Sq. l=1. fdl = lim f q. Sq. l=1. dl = f lim. q. Sq. l=1. dl = f. S. q=1. dq , which exists by hypothesis.. S S S (dq + eq ) converges. Then (dq + eq ) and dq are convergent series. So by Theorem 8, S S S S eq , a contradiction, since eq is given to be [(dq + eq ) dq ] would also be convergent. But [(dq + eq ) dq ] =. 69. Suppose on the contrary that. divergent. 71. The partial sums {vq } form an increasing sequence, since vq vq1 = dq A 0 for all q. Also, the sequence {vq } is bounded. since vq 1000 for all q. So by the Monotonic Sequence Theorem, the sequence of partial sums converges, that is, the series S dq is convergent..

(14) F. 466. ¤. TX.10. INFINITE SEQUENCES AND SERIES. CHAPTER 11. (length 13 ) is removed. At the second step, we remove the intervals 19 > 29 and 7 8 2 3 > , which have a total length of 2 · 13 . At the third step, we remove 22 intervals, each of length 13 . In general, 9 9 q q q1 . Thus, the total at the nth step we remove 2q1 intervals, each of length 13 , for a length of 2q1 · 13 = 13 23. 73. (a) At the rst step, only the interval. length of all removed intervals is. 1. 2 3> 3. S q=1. 1 3. . 2 q1. =. 3. 1@3 1 2@3. = 1 geometric series with d =. 1 3. and u =. 2 3. . Notice that at. q q , so we never remove 0, and 0 is in the Cantor set. Also, the qth step, the leftmost interval that is removed is 13 > 23 1 q 2 q , so 1 is never removed. Some other numbers in the Cantor set the rightmost interval removed is 1 3 > 1 3 are 13 , 23 , 19 , 29 , 79 , and 89 . 2 3 (b) The area removed at the rst step is 19 ; at the second step, 8 · 19 ; at the third step, (8)2 · 19 . In general, the area q q1 removed at the qth step is (8)q1 19 = 19 89 , so the total area of all removed squares is q1 1 S 1@9 8 = 1. = 9 1 8@9 q=1 9 75. (a) For. S. q 1 2 5 3 23 1 1 5 , v1 = = , v2 = + = , v3 = + = , (q + 1)! 1 · 2 2 2 1 · 2 · 3 6 6 1 · 2 · 3 · 4 24 q=1. v4 =. 4 119 23 (q + 1)! 1 + = . The denominators are (q + 1)!, so a guess would be vq = . 24 1·2·3·4·5 120 (q + 1)!. (b) For q = 1, v1 =. 2! 1 1 (n + 1)! 1 = , so the formula holds for q = 1. Assume vn = . Then 2 2! (n + 1)!. vn+1 = =. (n + 1)! 1 (n + 2)! (n + 2) + n + 1 (n + 1)! 1 n+1 n+1 + = + = (n + 1)! (n + 2)! (n + 1)! (n + 1)!(n + 2) (n + 2)! (n + 2)! 1 (n + 2)!. Thus, the formula is true for q = n + 1. So by induction, the guess is correct. S 1 (q + 1)! 1 q = lim 1 = 1 and so = 1. q q (q + 1)! (q + 1)! (q + 1)! q=1. (c) lim vq = lim q. 11.3 The Integral Test and Estimates of Sums 1. The picture shows that d2 =. d3 =. 1 31=3. ] ?. 2. 3. 1 {1=3. 1 21=3. ] ?. 2. 1. g{, and so on, so. 1 g{, {1=3 S q=2. 1 ? q1=3. ] 1. . 1 g{. The {1=3. integral converges by (7.8.2) with s = 1=3 A 1, so the series converges.. 5. 3. The function i ({) = 1@ { = {1@5 is continuous, positive, and decreasing on [1> ), so the Integral Test applies.. U 1. {1@5 g{ = lim. Uw. w 1. {1@5 g{ = lim. w. k. 5 4@5 { 4. lw 1. = lim. w. . 5 4@5 w 4. . 5 4. . = , so. S q=1. 1@ 5 q diverges..

(15) F. TX.10SECTION 11.3 THE INTEGRAL TEST AND ESTIMATES OF SUMS. ¤. 467. 1 is continuous, positive, and decreasing on [1> ), so the Integral Test applies. (2{ + 1)3 w ] w 1 1 1 1 1 1 1 . g{ = lim g{ = lim = lim + = w 1 (2{ + 1)3 w (2{ + 1)3 4 (2{ + 1)2 1 w 4(2w + 1)2 36 36. 5. The function i ({) =. ]. . 1. Since this improper integral is convergent, the series. S. 1 is also convergent by the Integral Test. (2q + 1)3 q=1. 7. i ({) = {h{ is continuous and positive on [1> ). i 0 ({) = {h{ + h{ = h{ (1 {) ? 0 for { A 1, so i is decreasing. on [1> ). Thus, the Integral Test applies. U { Ue e {h g{ = lim 1 {h{ g{ = lim {h{ h{ 1 1 e. e. [by parts] = lim [ehe he + h1 + h1 ] = 2@h. since lim ehe = lim (e@he ) = lim (1@he ) = 0 and lim he = 0. Thus, H. e. 9. The series. e. S q=1. e. e. S. e. q=1. qhq converges.. S 1 2 is a s-series with s = 0=85 1, so it diverges by (1). Therefore, the series must also diverge, 0=85 q0=85 q q=1. for if it converged, then. S. 1 would have to converge [by Theorem 8(i) in Section 11.2]. 0=85 q q=1. 11. 1 +. 1 S 1 1 1 1 + + + +··· = . This is a s-series with s = 3 A 1, so it converges by (1). 3 8 27 64 125 q=1 q. 13. 1 +. S 1 1 1 1 1 1 + + + + ··· = . The function i({) = is 3 5 7 9 2q 1 2{ 1 q=1. continuous, positive, and decreasing on [1> ), so the Integral Test applies. ] ] w w S 1 1 1 g{ = lim g{ = lim 12 ln |2{ 1| 1 = 12 lim (ln(2w 1) 0) = , so the series w w w 2{ 1 2{ 1 2q 1 q=1 1 1 diverges. 15.. 52 1 1 S S S S S q 1 1 =5 2 by Theorem 11.2.8, since and both converge by (1) 3 3 3 5@2 5@2 q q=1 q=1 q q=1 q q=1 q q=1 q 52 S q 5 with s = 3 A 1 and s = 2 A 1 . Thus, converges. q3 q=1 1 is continuous, positive, and decreasing on [1> ), so we can apply the Integral Test. {2 + 4 w ] w 1 1 1 w 1 1 { 1 1 1 tan g{ = lim g{ = lim tan lim tan = w 1 {2 + 4 w 2 {2 + 4 2 1 2 w 2 2 1 1 = tan1 2 2 2. 17. The function i ({) =. ] 1. . Therefore, the series. S q=1. 19.. 1 converges. q2 + 4. ln q ln q S S ln { ln 1 = 0. The function i ({) = 3 is continuous and positive on [2> ). = since 3 3 q q 1 { q=1 q=2. i 0 ({) =. {3 (1@{) (ln {)(3{2 ) {2 3{2 ln { 1 3 ln { = = ? 0 1 3 ln { ? 0 ln { A 3 2 ({ ) {6 {4. 1 3.

(16) F. 468. ¤. INFINITE SEQUENCES AND SERIES. CHAPTER 11. TX.10. { A h1@3 1=4, so i is decreasing on [2> ), and the Integral Test applies. w ] w ] ln q S (B) 1 1 (BB) 1 ln { ln { ln { 1 g{ = lim g{ = lim = lim (2 ln w + 1) + = , so the series 3 3 2 2 2 3 w w w { { 2{ 4{ 1 4w 4 4 q=2 q 2 2 converges. (B): x = ln {, gy = {3 g{ gx = (1@{) g{, y = 12 {2 , so ] ] ] ln { 1 2 1 2 1 2 1 {3 g{ = 12 {2 ln { 14 {2 + F= g{ = { ln { { (1@{) g{ = { ln { + 2 2 2 2 {3 2 ln w + 1 H 2@w 1 = 14 lim 2 = 0. = lim (BB): lim w w 8w w w 4w2 1 1 + ln { is continuous and positive on [2> ), and also decreasing since i 0 ({) = 2 ? 0 for { A 2, so we can { ln { { (ln {)2 ] S 1 1 g{ = lim [ln(ln {)]w2 = lim [ln(ln w) ln(ln 2)] = , so the series diverges. use the Integral Test. w w { ln { q ln q q=2 2. 21. i ({) =. 23. The function i ({) = h1@{@{2 is continuous, positive, and decreasing on [1> ), so the Integral Test applies.. [j({) = h1@{ is decreasing and dividing by {2 doesn’t change that fact.] ] w 1@{ ] k lw h1@q S h i ({) g{ = lim g{ = lim h1@{ = lim (h1@w h) = (1 h) = h 1, so the series 2 2 w 1 w w { 1 q=1 q 1 converges. 25. The function i ({) =. 1 is continuous, positive, and decreasing on [1> ), so the Integral Test applies. We use partial {3 + {. fractions to evaluate the integral: w w ] ] w { 1 1 1 { 2 g{ = lim ln(1 + { ) = lim g{ = lim ln { ln w 1 w w {3 + { { 1 + {2 2 1 + {2 1 1 1 $ # 1 1 1 w 1 = lim ln ln + ln 2 = ln 2 = lim ln s w w 2 2 1 + w2 2 1 + 1@w2 so the series. S. 1 converges. 3 +q q q=1. 27. We have already shown (in Exercise 21) that when s = 1 the series. i ({) =. S q=2. 1 diverges, so assume that s 6= 1. q(ln q)s. 1 s + ln { is continuous and positive on [2> ), and i 0 ({) = 2 ? 0 if { A hs , so that i is eventually {(ln {)s { (ln {)s+1. decreasing and we can use the Integral Test. ] 2. . w (ln {)1s 1 g{ = lim w {(ln {)s 1s 2. [for s 6= 1] = lim. w. (ln w)1s (ln 2)1s 1s 1s. This limit exists whenever 1 s ? 0 s A 1, so the series converges for s A 1.. .

(17) F. TX.10SECTION 11.3 THE INTEGRAL TEST AND ESTIMATES OF SUMS. ¤. 469. 29. Clearly the series cannot converge if s 12 , because then lim q(1 + q2 )s 6= 0. So assume s ? 12 . Then q. i ({) = {(1 + {2 )s is continuous, positive, and eventually decreasing on [1> ), and we can use the Integral Test. ]. . . 2 s. {(1 + { ) g{ = lim. w. 1. 1 (1 + {2 )s+1 · 2 s+1. w = 1. 1 lim [(1 + w2 )s+1 2s+1 ]. 2(s + 1) w. This limit exists and is nite s + 1 ? 0 s ? 1, so the series converges whenever s ? 1. 31. Since this is a s-series with s = {, ({) is dened when { A 1. Unless specied otherwise, the domain of a function i is the. set of real numbers { such that the expression for i({) makes sense and denes a real number. So, in the case of a series, it’s the set of real numbers { such that the series is convergent. 1 2 is positive and continuous and i 0 ({) = 3 is negative for { A 0, and so the Integral Test applies. {2 {. 33. (a) i ({) =. 1 S 1 1 1 1 v10 = 2 + 2 + 2 + · · · + 2 1=549768. 2 1 2 3 10 q=1 q w ] 1 1 1 1 1 + = , so the error is at most 0=1. g{ = lim = lim U10 2 w { 10 w w 10 10 10 {. ] (b) v10 +. . 11. 1 g{ v v10 + {2. ]. . 10. 1 g{ v10 + {2. 1 11. v v10 +. 1 10. . 1=549768 + 0=090909 = 1=640677 v 1=549768 + 0=1 = 1=649768, so we get v 1=64522 (the average of 1=640677 and 1=649768) with error 0=005 (the maximum of 1=649768 1=64522 and 1=64522 1=640677, rounded up). ] 1 1 1 1 q A 1000. g{ = . So Uq ? 0=001 if ? (c) Uq 2 { q q 1000 q 35. i ({) = 1@(2{ + 1)6 is continuous, positive, and decreasing on [1> ), so the Integral Test applies. Using (2),. ] Uq . . q. (2{ + 1)6 g{ = lim. w. 5 1 6 10(2q + 1)5 10 v4 =. 37.. S. 4 S q=1. . 1 10(2{ + 1)5. w q. =. 1 . To be correct to ve decimal places, we want 10(2q + 1)5. (2q + 1)5 20,000 q . 1 2. 5 20,000 1 3=12, so use q = 4.. 1 1 1 1 1 = 6 + 6 + 6 + 6 0=001 446 0=00145. (2q + 1)6 3 5 7 9 S. 1 is a convergent s-series with s = 1=001 A 1. Using (2), we get q1=001 0=001 w w ] { 1 1 1000 {1=001 g{ = lim = 1000 lim = 1000 Uq = 0=001 . 0=001 0=001 w w 0=001 { q q q q q. q=1. q1=001 =. q=1. We want Uq ? 0=000 000 005 . 1000 ? 5 × 109 q0=001. q0=001 A. 1000 5 × 109. 1000 q A 2 × 1011 = 21000 × 1011,000 1=07 × 10301 × 1011,000 = 1=07 × 1011,301 ..

(18) F. 470. ¤. 39. (a) From the gure, d2 + d3 + · · · + dq . i ({) =. TX.10. INFINITE SEQUENCES AND SERIES. CHAPTER 11. 1 1 1 1 1 , + + + ··· + { 2 3 4 q. Uq 1. ]. q. 1. i ({) g{, so with 1 g{ = ln q. {. 1 1 1 1 Thus, vq = 1 + + + + · · · + 1 + ln q. 2 3 4 q (b) By part (a), v106 1 + ln 106 14=82 ? 15 and v109 1 + ln 109 21=72 ? 22. . 41. eln q = hln e. ln q. ln e = hln q = qln e =. ln e ? 1 e ? h1. 1 . This is a s-series, which converges for all e such that ln e A 1 q ln e. e ? 1@h [with e A 0].. 11.4 The Comparison Tests 1. (a) We cannot say anything about. S. dq . If dq A eq for all q and. S. eq is convergent, then. S. dq could be convergent or. divergent. (See the note after Example 2.) S (b) If dq ? eq for all q, then dq is convergent. [This is part (i) of the Comparison Test.] 3.. 1 S S q q q 1 1 ? converges by comparison with = ? 2 for all q 1, so , which converges 3 2 3 2 +1 2q 2q q q=1 2q + 1 q=1 q. 2q3. because it is a p-series with s = 2 A 1. 5.. q+1 S S q+1 1 q 1 A = for all q 1, so diverges by comparison with , which diverges because it is a q=1 q q=1 q q q q q q q. p-series with s = 7.. 1 2. 9q 9q ? q = q 3 + 10 10. 1. . 9 10. q. for all q 1.. S q=1. 9 q 10. is a convergent geometric series |u| =. 9 10. S ? 1 , so. q=1. 9q 3 + 10q. converges by the Comparison Test. 9.. 11.. 13.. cos2 q 1 S S cos2 q 1 1 2 ? 2 , so the series converges by comparison with the s-series 2 2 2 q +1 q +1 q q=1 q + 1 q=1 q. q1 q1 q 1 is positive for q A 1 and ? = q = q 4q q 4q q 4q 4 q S 1 . geometric series 4 q=1. [s = 2 A 1].. q q1 S 1 , so converges by comparison with the convergent q 4 q=1 q 4. arctan q S arctan q 1 @2 S ? 1=2 for all q 1, so converges by comparison with , which converges because it is a 1=2 1=2 q1=2 q q 2 q q=1 q=1. constant times a p-series with s = 1=2 A 1. 15.. S S 2 + (1)q 3 1 3 converges because it is a constant multiple of the convergent s-series , and q q q q q q q q q=1 q=1 s = 32 A 1 , so the given series converges by the Comparison Test..

(19) F. TX.10 17. Use the Limit Comparison Test with dq = . 1. q2. +1. and eq =. SECTION 11.4. THE COMPARISON TESTS. ¤. 471. 1 : q. 1 S dq q 1 = 1 A 0. Since the harmonic series diverges, so does = lim = lim s lim 2 2 q eq q q q +1 1 + (1@q ) q=1 q S. 1 . q2 + 1. q=1. 19. Use the Limit Comparison Test with dq =. 1 + 4q 4q and eq = q : 1 + 3q 3. 1 + 4q q 1 dq 1 + 4q 3q 1 + 4q 3q 1 =1A0 = lim 1 +q3 = lim · = lim · = lim + 1 · lim 1 q eq q 4 q 1 + 3q q q 4q 4q 1 + 3q 4q + 1 3q 3q q 1+4 S S 4 q S Since the geometric series eq = diverges, so does . Alternatively, use the Comparison Test with 3 q q=1 1 + 3 q 1 + 4q 1 4 1 + 4q 4q = A A or use the Test for Divergence. 1 + 3q 3q + 3q 2(3q ) 2 3 q+2 1 and eq = 3@2 : 2q2 + q + 1 q s 1 + 2@q dq q3@2 q + 2 (q3@2 q + 2 )@(q3@2 q ) 1 1 = lim = A 0. = lim = lim = lim q eq q 2q2 + q + 1 q q 2 + 1@q + 1@q2 (2q2 + q + 1)@q2 2 2 S S 1 q+2 Since also converges. is a convergent p-series s = 32 A 1 , the series 2 +q+1 3@2 2q q q=1 q=1. 21. Use the Limit Comparison Test with dq =. 23. Use the Limit Comparison Test with dq =. 5 + 2q 1 and eq = 3 : (1 + q2 )2 q. dq q3 (5 + 2q) 5q3 + 2q4 1@q4 = lim = lim · = lim q eq q (1 + q2 )2 q (1 + q2 )2 q 1@(q2 )2 lim. s-series [s = 3 A 1], the series. S q=1. 5 q 1 q2. 1 S is a convergent 2 = 2 A 0. Since 3 q=1 q +1. +2. 5 + 2q also converges. (1 + q2 )2. dq q + q2 + q3 1@q2 + 1@q + 1 1 + q + q2 1 = 1 A 0, and eq = , then lim = lim = lim s 2 6 2 6 q eq q q q 1+q +q 1+q +q 1@q6 + 1@q4 + 1. 25. If dq = S. so. q=1. 1 S 1 + q + q2 . diverges by the Limit Comparison Test with the divergent harmonic series 2 6 1+q +q q=1 q. 2 2 1 1 dq 1+ hq and eq = hq : lim = lim 1 + = 1 A 0. Since q eq q q q 2 1 S S 1 1 = is a convergent geometric series |u| = ? 1 , the series hq also converges. 1 + h q q q=1 h q=1. 27. Use the Limit Comparison Test with dq = S q=1. hq. 29. Clearly q! = q(q 1)(q 2) · · · (3)(2) 2 · 2 · 2 · · · · · 2 · 2 = 2q1 , so. series |u| =. 1 2. 1 S converges by the Comparison Test. ? 1 , so q=1 q!. S 1 1 1 q1 . is a convergent geometric q1 q! 2 q=1 2.

(20) F. 472. ¤. CHAPTER 11. INFINITE SEQUENCES AND SERIES. TX.10. S S 1 1 31. Use the Limit Comparison Test with dq = sin and eq = . Then dq and eq are series with positive terms and q q S dq sin(1@q) sin = lim = 1 A 0. Since = lim eq is the divergent harmonic series, q eq q 0 1@q q=1. lim. S q=1. sin (1@q) also diverges. [Note that we could also use l’Hospital’s Rule to evaluate the limit:. lim. {. 33.. 35.. cos(1@{) · 1@{2 sin(1@{) H 1 = lim = lim cos = cos 0 = 1.] { { 1@{ 1@{2 {. 10 S. 1 1 1 1 1 1 1 1 1=24856. Now = + + + ··· + ? = 2 , so the error is 4 4 4 q 10,001 q +1 q +1 2 17 82 q q=1 w ] 1 1 1 1 1 = = 0=1. g{ = lim = lim + U10 W10 2 w w { { w 10 10 10 10 10 S q=1. 1 1 1 1 1 1 1 0=76352. Now = + + + ··· + ? q , so the error is 1 + 2q 3 5 9 1025 1 + 2q 2. U10 W10 =. 37. Since. S q=11. 1@211 1 = 2q 1 1@2. [geometric series] 0=00098.. S 9 gq 9 q for each q, and since is a convergent geometric series |u| = q q 10 10 q=1 10. 1 10. g S q ? 1 , 0=g1 g2 g3 = = = = q q=1 10. will always converge by the Comparison Test. 39. Since. S. dq converges, lim dq = 0, so there exists Q such that |dq 0| ? 1 for all q A Q q. all q A Q. 0 d2q dq . Since. S. dq converges, so does. S. 0 dq ? 1 for. d2q by the Comparison Test.. dq dq = , there is an integer Q such that A 1 whenever q A Q. (Take P = 1 in Denition 11.1.5.) eq eq S S Then dq A eq whenever q A Q and since eq is divergent, dq is also divergent by the Comparison Test.. 41. (a) Since lim. q. (b) (i) If dq =. 1 1 dq q { H 1 and eq = for q 2, then lim = lim = lim { = , = lim = lim q eq q ln q { ln { { 1@{ { ln q q. so by part (a),. S q=2. (ii) If dq = so. S q=1. S dq ln q 1 and eq = , then eq is the divergent harmonic series and lim = lim ln q = lim ln { = , q eq q { q q q=1. dq diverges by part (a).. 43. lim qdq = lim q. 1 is divergent. ln q. q. dq 1 , so we apply the Limit Comparison Test with eq = . Since lim qdq A 0 we know that either both q 1@q q. series converge or both series diverge, and we also know that divergent.. 1 S S diverges [s-series with s = 1]. Therefore, dq must be q=1 q.

(21) F. TX.10 45. Yes. Since. S. SECTION 11.5. ALTERNATING SERIES. dq is a convergent series with positive terms, lim dq = 0 by Theorem 11.2.6, and q. series with positive terms (for large enough q). We have lim. q. S. eq =. S. ¤. 473. sin(dq ) is a. S eq sin(dq ) = lim = 1 A 0 by Theorem 3.3.2. Thus, eq q dq dq. is also convergent by the Limit Comparison Test.. 11.5 Alternating Series 1. (a) An alternating series is a series whose terms are alternately positive and negative.. (b) An alternating series. S. (1)q1 eq converges if 0 ? eq+1 eq for all q and lim eq = 0. (This is the Alternating. q=1. q. Series Test.) (c) The error involved in using the partial sum vq as an approximation to the total sum v is the remainder Uq = v vq and the size of the error is smaller than eq+1 ; that is, |Uq | eq+1 . (This is the Alternating Series Estimation Theorem.) 3.. S 4 4 4 4 4 4 4 + + ··· = . Now eq = A 0, {eq } is decreasing, and lim eq = 0, so the (1)q1 q 7 8 9 10 11 q+6 q+6 q=1. series converges by the Alternating Series Test. 5.. S q=1. dq =. S q=1. (1)q1. S 1 1 = A 0, {eq } is decreasing, and lim eq = 0, so the (1)q1 eq . Now eq = q 2q + 1 q=1 2q + 1. series converges by the Alternating Series Test. 7.. S q=1. dq =. S. (1)q. q=1. S 3q 1 3 1@q 3 = = 6= 0. Since lim dq 6= 0 (1)q eq . Now lim eq = lim q q 2 + 1@q q 2q + 1 q=1 2. (in fact the limit does not exist), the series diverges by the Test for Divergence. q A 0 for q 1. {eq } is decreasing for q 1 since 10q { 0 10{ (1) { · 10{ ln 10 10{ (1 { ln 10) 1 { ln 10 = = = ? 0 for 1 { ln 10 ? 0 { ln 10 A 1 10{ (10{ )2 (10{ )2 10{. 9. eq =. {A. S q { H { q 1 0=4. Also, lim eq = lim = 0. Thus, the series = lim = lim (1)q q q q 10q { 10{ { 10{ ln 10 ln 10 10 q=1. converges by the Alternating Series Test. q2 A 0 for q 1. {eq } is decreasing for q 2 since q3 + 4 0 ({3 + 4)(2{) {2 (3{2 ) {(2{3 + 8 3{3 ) {(8 {3 ) {2 = = = 3 ? 0 for { A 2. Also, 3 3 2 3 2 { +4 ({ + 4) ({ + 4) ({ + 4)2. 11. eq =. [ 1@q q2 q+1 converges by the Alternating Series Test. = 0. Thus, the series (1) q 1 + 4@q3 q3 + 4 q=1. lim eq = lim. q. 13.. S. (1)q. q=2. q q { H 1 . lim = lim = , so the series diverges by the Test for Divergence. = lim ln q q ln q { ln { { 1@{.

(22) F. 474. ¤. CHAPTER 11. TX.10. INFINITE SEQUENCES AND SERIES q. 15.. cos q (1) S S 1 1 = . eq = 3@4 is decreasing and positive and lim 3@4 = 0, so the series converges by the 3@4 3@4 q q q q q q=1 q=1. Alternating Series Test. . eq = sin A 0 for q 2 and sin sin , and lim sin = sin 0 = 0, so the series 17. (1) sin q q q q q+1 q q=1 S. q. converges by the Alternating Series Test. 19.. qq q· q · ··· · q = q q! 1 · 2 · ··· · q. qq = q q! lim. lim. q. (1)q qq does not exist. So the series diverges by the Test for q!. Divergence. 21.. q 1 2 3 4. dq 1. vq 1. 0=35355. 0=64645. 0=19245. 0=83890. 0=125. 0=71390. 5. 0=08944. 0=80334. 6. 0=06804. 0=73530. 7. 0=05399. 0=78929. 8. 0=04419. 0=74510. 9. 0=03704. 0=78214. 10. 0=03162. 0=75051. 23. The series. By the Alternating Series Estimation Theorem, the error in the approximation (1)q1 S 0=75051 is |v v10 | e11 = 1@(11)3@2 0=0275 (to four q3@2 q=1. decimal places, rounded up).. (1)q+1 S 1 1 1 satises (i) of the Alternating Series Test because ? 6 and (ii) lim 6 = 0, so the q q q6 (q + 1)6 q q=1. series is convergent. Now e5 =. 1 1 = 0=000064 A 0=00005 and e6 = 6 0=00002 ? 0=00005, so by the Alternating Series 56 6. Estimation Theorem, q = 5. (That is, since the 6th term is less than the desired error, we need to add the rst 5 terms to get the sum to the desired accuracy.) 25. The series. (1)q S 1 1 1 satises (i) of the Alternating Series Test because q+1 ? q and (ii) lim = 0, q q 10q q! 10 (q + 1)! 10 q! q=0 10 q!. so the series is convergent. Now e3 =. 1 1 0=000 167 A 0=000 005 and e4 = 4 = 0=000 004 ? 0=000 005, so by 103 3! 10 4!. the Alternating Series Estimation Theorem, q = 4 (since the series starts with q = 0, not q = 1). (That is, since the 5th term is less than the desired error, we need to add the rst 4 terms to get the sum to the desired accuracy.) 27. e7 =. 1 1 0=000 059 5, so = 75 16,807. (1)q+1 6 (1)q+1 S S. v = = 1 6 q5 q5 q=1 q=1. 1 32. +. 1 243. . 1 1024. +. 1 3125. . 1 7776. 0=972 080. Adding e7 to v6 does not change. the fourth decimal place of v6 , so the sum of the series, correct to four decimal places, is 0=9721..

(23) F. TX.10 SECTION ON N 11.6 ABSOLUTE CONVERGENCE AND THE RATIO AND ROOT TESTS 29. e7 =. ¤. 475. 72 = 0=000 004 9, so 107. (1)q1 q2 6 (1)q1 q2 S S v6 = = q 10 10q q=1 q=1. 1 10. . 4 100. +. 9 1000. . 16 10,000. +. 25 100,000. . 36 1,000,000. = 0=067 614. Adding e7 to v6. does not change the fourth decimal place of v6 , so the sum of the series, correct to four decimal places, is 0=0676. 31.. (1)q1 S 1 1 1 1 1 1 1 = 1 + + ··· + + + · · · . The 50th partial sum of this series is an q 2 3 4 49 50 51 52 q=1 (1)q1 S 1 1 1 1 underestimate, since = v50 + + + · · · , and the terms in parentheses are all positive. q 51 52 53 54 q=1. The result can be seen geometrically in Figure 1. 33. Clearly eq =. 1 is decreasing and eventually positive and lim eq = 0 for any s. So the series converges (by the q q+s. Alternating Series Test) for any s for which every eq is dened, that is, q + s 6= 0 for q 1, or s is not a negative integer. 35.. S. e2q =. S. 1@(2q)2 clearly converges (by comparison with the s-series for s = 2). So suppose that. converges. Then by Theorem 11.2.8(ii), so does. S (1)q1 eq + eq = 2 1 +. diverges by comparison with the harmonic series, a contradiction. Therefore,. S. 1 3. +. 1 5. S. S +··· = 2. (1)q1 eq 1 . But this 2q 1. (1)q1 eq must diverge. The Alternating. Series Test does not apply since {eq } is not decreasing.. 11.6 Absolute Convergence and the Ratio and Root Tests . . dq+1 = 8 A 1, part (b) of the Ratio Test tells us that the series 1. (a) Since lim q dq . S. dq is divergent.. dq+1 = 0=8 ? 1, part (a) of the Ratio Test tells us that the series S dq is absolutely convergent (and (b) Since lim q dq therefore convergent). dq+1 = 1, the Ratio Test fails and the series S dq might converge or it might diverge. (c) Since lim q dq 3.. q+1 (10)q dq+1 10 S q! = 0 ? 1, so the series is = lim (10) . Using the Ratio Test, lim · = lim q q! dq q (q + 1)! (10)q q q + 1 q=0 absolutely convergent.. 5.. [ [ (1)q+1 1 converges by the Alternating Series Test, but is a divergent s-series s = 4 4 q q q=1 q=1. is conditionally convergent.. 1 4. 1 , so the given series.

(24) F. ¤. 476. . TX.10. INFINITE SEQUENCES AND SERIES. CHAPTER 11. . dn+1 = lim 7. lim n dn n. %. 2 n+1 &. (n + 1) 3 n n 23. = lim. n. n+1 n. 1 2 1 2 lim 1 + = = 23 (1) = 3 3 n n. 2 3. ? 1, so the series. n S n 23 is absolutely convergent by the Ratio Test. Since the terms of this series are positive, absolute convergence is the. q=1. same as convergence. q+1 dq+1 (1=1)q4 1 1 q4 = lim (1=1) 9. lim · = (1=1) lim = (1=1) lim = lim 4 q q (q + 1) q (q + 1)4 q (q + 1)4 q (1 + 1@q)4 dq (1=1)q q4 = (1=1)(1) = 1=1 A 1, S. so the series. (1)q. q=1. 11. Since 0 . (1=1)q diverges by the Ratio Test. q4. 1 h1@q S S h1@q h 1 and = h is a convergent s-series [s = 3 A 1], converges, and so 3 3 3 3 3 q q q q=1 q q=1 q. (1)q h1@q S is absolutely convergent. q3 q=1. . . . dq+1 10q+1 (q + 1) 42q+1 = lim 13. lim · 2q+3 q q (q + 2) 4 dq 10q. . = lim. q. 10 q + 1 · 42 q + 2. =. S 10q 5 ? 1, so the series 2q+1 8 q=1 (q + 1)4. is absolutely convergent by the Ratio Test. Since the terms of this series are positive, absolute convergence is the same as convergence. . . @2 1 (1)q arctan q (1)q arctan q @2 S S S ? 15. , so since = converges (s = 2 A 1), the given series 2 2 2 2 2. q. q. q=1. q. 2. q=1. q. q=1. q. converges absolutely by the Comparison Test. 17.. (1)q S 1 converges by the Alternating Series Test since lim = 0 and q ln q q=2 ln q. . 1 ln q. is decreasing. Now ln q ? q, so. 1 S S 1 1 1 A , and since is the divergent (partial) harmonic series, diverges by the Comparison Test. Thus, ln q q q ln q q=2 q=2 (1)q S is conditionally convergent. q=2 ln q. 19.. 1 cos(q@3) S S |cos (q@3)| 1 and converges (use the Ratio Test), so the series converges absolutely by the q! q! q! q=1 q! q=1. Comparison Test. 21. lim. q. s S 1 q2 + 1 1 + 1@q2 q = lim |dq | = lim = ? 1, so the series 2 2 q 2q + 1 q 2 + 1@q 2 q=1. . q2 + 1 2q2 + 1. q is absolutely convergent by the. Root Test. s 23. lim q |dq | = lim q. q. v q2 q q2 S 1 1 1 q 1+ 1+ = lim 1 + = h A 1 (by Equation 3.6.6), so the series q q q q q=1. diverges by the Root Test..

(25) F. TX.10 SECTION ON N 11.6 ABSOLUTE CONVERGENCE AND THE RATIO AND ROOT TESTS. ¤. 477. 25. Use the Ratio Test with the series. 1. S 1 · 3 · 5 · · · · · (2q 1) 1 · 3 · 5 · · · · · (2q 1) 1·3·5 1·3·5·7 1·3 + + · · · + (1)q1 + ··· = . (1)q1 3! 5! 7! (2q 1)! (2q 1)! q=1 dq+1 (1)q · 1 · 3 · 5 · · · · · (2q 1)[2(q + 1) 1] (2q 1)! lim · = lim q dq q [2(q + 1) 1]! (1)q1 · 1 · 3 · 5 · · · · · (2q 1) (1)(2q + 1)(2q 1)! = lim 1 = 0 ? 1, = lim q (2q + 1)(2q)(2q 1)! q 2q. so the given series is absolutely convergent and therefore convergent. 27.. 2 · 4 · 6 · · · · · (2q) (2 · 1) · (2 · 2) · (2 · 3) · · · · · (2 · q) 2q q! S S S S = = = 2q , which diverges by the Test for q! q! q! q=1 q=1 q=1 q=1. Divergence since lim 2q = . q. dq+1 5q + 1 5 = A 1, so the series diverges by the Ratio Test. = lim 29. By the recursive denition, lim q dq q 4q + 3 4 1@(q + 1)3 q3 1 = lim 31. (a) lim = lim = 1. Inconclusive 3 q q (q + 1)3 q (1 + 1@q)3 1@q (q + 1) 2q 1 q+1 1 1 = lim = lim + (b) lim q+1 · = . Conclusive (convergent) q q 2 2 q q 2q 2q 2 u u (3)q q q 1 = 3 lim = 3. Conclusive (divergent) = 3 lim (c) lim · q1 q q q q+1 1 + 1@q q + 1 (3) %u & 2 2 1@q + 1 1 + q 1 q + 1 (d) lim · = lim 1+ · = 1. Inconclusive q 1 + (q + 1)2 q q 1@q2 + (1 + 1@q)2 q q+1 dq+1 { { q! 1 = |{| lim · = |{| · 0 = 0 ? 1, so by the Ratio Test the = lim = lim 33. (a) lim q q q + 1 dq q (q + 1)! {q q q + 1 series. {q S converges for all {. q=0 q!. (b) Since the series of part (a) always converges, we must have lim. q. 35. (a) v5 =. uq =. {q = 0 by Theorem 11.2.6. q!. 5 S. 1 1 1 1 661 1 1 + + = 0=68854. Now the ratios = + + q q2 2 8 24 64 160 960 q=1 dq+1 q2q q form an increasing sequence, since = = dq (q + 1)2q+1 2(q + 1). q (q + 1)2 q(q + 2) 1 q+1 = = A 0. So by Exercise 34(b), the error 2(q + 2) 2(q + 1) 2(q + 1)(q + 2) 2(q + 1)(q + 2) 1@ 6 · 26 1 d6 = 0=00521. = in using v5 is U5 1 lim uq 1 1@2 192. uq+1 uq =. q. (b) The error in using vq as an approximation to the sum is Uq =. dq+1 2 = . We want Uq ? 0=00005 (q + 1)2q+1 1 12. 1 ? 0=00005 (q + 1)2q A 20,000. To nd such an q we can use trial and error or a graph. We calculate (q + 1)2q (11 + 1)211 = 24,576, so v11 =. 11 S. 1 0=693109 is within 0=00005 of the actual sum. q q2 q=1.

(26) F. 478. ¤. INFINITE SEQUENCES AND SERIES. CHAPTER 11. TX.10. 37. (i) Following the hint, we get that |dq | ? uq for q Q, and so since the geometric series. S. uq converges [0 ? u ? 1], S S S the series q=Q |dq | converges as well by the Comparison Test, and hence so does q=1 |dq |, so q=1 dq is absolutely q=1. convergent. (ii) If lim. q. s s q |dq | = O A 1, then there is an integer Q such that q |dq | A 1 for all q Q, so |dq | A 1 for q Q. Thus,. lim dq 6= 0, so. S. q=1. q. (iii) Consider. dq diverges by the Test for Divergence.. 1 1 s S S [diverges] and [converges]. For each sum, lim q |dq | = 1, so the Root Test is inconclusive. 2 q q q q=1 q=1. + dq is absolutely convergent, and since d+ q |dq | and dq |dq | (because dq and dq each equal S S either dq or 0), we conclude by the Comparison Test that both d+ dq must be absolutely convergent. q and. 39. (a) Since. S. Or: Use Theorem 11.2.8. S S S (b) We will show by contradiction that both d+ dq must diverge. For suppose that d+ q and q converged. Then so S + 1 S 1 S S + 1 1 dq 2 dq = dq 2 dq by Theorem 11.2.8. But = 12 |dq |, which would 2 (dq + |dq |) 2 dq S + S S diverges because dq is only conditionally convergent. Hence, dq can’t converge. Similarly, neither can d q.. 11.7 Strategy for Testing Series 1.. 1 1 ? q = q + 3q 3. q 1 for all q 1. 3. q 1 is a convergent geometric series |u| = 3 q=1 S. 1 3. S ? 1 , so. q=1. 1 q + 3q. converges by the Comparison Test. S q q q = 1, so lim dq = lim (1)q does not exist. Thus, the series diverges by (1)q q q + 2 q q q+2 q+2 q=1. 3. lim |dq | = lim q. the Test for Divergence. . . . . . 2 q q 2 dq+1 = lim (q + 1) 2 · (5) = lim 2(q + 1) = 2 lim 1 + 1 5. lim q+1 2 q1 2 q q q dq q 2 5q 5 q q (5). 2. =. 2 2 (1) = ? 1, so the series 5 5. q2 2q1 S q converges by the Ratio Test. q=1 (5). 1 . Then i is positive, continuous, and decreasing on [2> ), so we can apply the Integral Test. ln { % & ] ] x = ln {, 1 g{ = x1@2 gx = 2x1@2 + F = 2 ln { + F, we nd Since { ln { gx = g{@{ ] w ] lw k g{ g{ = lim = lim 2 ln { = lim 2 ln w 2 ln 2 = . Since the integral diverges, the w 2 { ln { w 2 { ln { w 2. 7. Let i ({) =. given series. {. S q=2. q. 1 diverges. ln q.

(27) F. SECTION 11.7. 9.. S. STRATEGY FOR TESTING SERIES. ¤. 479. n2 S . Using the Ratio Test, we get n n=1 n=1 h % & 2 dn+1 (n + 1)2 hn 1 1 1 n + 1 = lim lim · 2 = lim · = 12 · = ? 1, so the series converges. n dn n hn+1 n n n h h h. n2 hn =. 11. eq =. (1)q+1 S 1 A 0 for q 2, {eq } is decreasing, and lim eq = 0, so the given series converges by the q q ln q q=2 q ln q. Alternating Series Test. q+1 3q q2 dq+1 3 S (q + 1)2 3(q + 1)2 q+1 q! · q 2 = lim = lim 13. lim = 3 lim = 0 ? 1, so the series q q q (q + 1)q2 q q2 dq (q + 1)! 3 q q=1 q! converges by the Ratio Test. . . . . dq+1 (q + 1)! q+1 2 · 5 · 8 · · · · · (3q + 2) 1 = lim · lim = ? 1, 15. lim = q q q 2 · 5 · 8 · · · · · (3q + 2)[3(q + 1) + 2] dq q! 3q + 5 3 so the series. S q=0. q! converges by the Ratio Test. 2 · 5 · 8 · · · · · (3q + 2) q. 17. lim 21@q = 20 = 1, so lim (1) 21@q does not exist and the series q. q. ln {. 19. Let i({) = . Then i 0 ({) =. {. S. (1)q 21@q diverges by the Test for Divergence.. q=1. 2 ln { ln q ? 0 when ln { A 2 or { A h2 , so is decreasing for q A h2 . 2{3@2 q. S ln q 1@q 2 ln q = lim = 0, so the series By l’Hospital’s Rule, lim = lim (1)q converges by the q q q q=1 q q q 1@ 2 q. Alternating Series Test. 21.. q (2)2q s S S 4 4 = 0 ? 1, so the given series is absolutely convergent by the Root Test. = . lim q |dq | = lim q q q q q q=1 q=1 q 1 1 and eq = , we have q q. 23. Using the Limit Comparison Test with dq = tan. tan(1@q) tan(1@{) H sec2 (1@{) · (1@{2 ) dq = lim = lim = lim = lim sec2 (1@{) = 12 = 1 A 0. Since q eq q { { { 1@q 1@{ 1@{2 lim. S q=1. eq is the divergent harmonic series,. S q=1. dq is also divergent.. 2 (q + 1)! hq2 q! dq+1 S (q + 1)q! · hq q+1 = lim = lim = lim 2q+1 = 0 ? 1, so 25. Use the Ratio Test. lim · 2 2 +2q+1 (q+1) q q2 q q h q h dq q! q h q! q=1 h converges. ] 27. 2. . w ln { ln { 1 g{ = lim w {2 { { 1. H. [using integration by parts] = 1. So. ln q S converges by the Integral Test, and since 2 q=1 q. S n ln n ln n n ln n n ln n = 2 , the given series 3 ? 3 converges by the Comparison Test. 3 n n (n + 1) n=1 (n + 1).

(28) F. 480. 29.. ¤ S q=1. CHAPTER 11. dq =. S. TX.10. INFINITE SEQUENCES AND SERIES. (1)q. q=1. S 1 1 = A 0, {eq } is decreasing, and lim eq = 0, so the series (1)q eq . Now eq = q cosh q q=1 cosh q. converges by the Alternating Series Test. Or: Write. 1 S S 1 1 2 2 = q is convergent by the ? q and is a convergent geometric series, so q q cosh q h +h h q=1 h q=1 cosh q S. Comparison Test. So. (1)q. q=1. 1 is absolutely convergent and therefore convergent. cosh q n n (5@4)n 3 5 = since lim = 0 and lim = . n (3@4)n + 1 n 4 n 4. 5n = [divide by 4n ] n n 3 + 4n. 31. lim dn = lim n. Thus,. S n=1. 33. Let dq =. 3n. 5n diverges by the Test for Divergence. + 4n. sin(1@q) S sin(1@q) sin(1@q) 1 dq = 1 A 0, so = lim and eq = . Then lim converges by limit q eq q 1@q q q q q q=1. comparison with the convergent s-series. 35. lim. q. lim. s q |dq | = lim. . q. q q+1. q2 @q. S. 1 q3@2. q=1. = lim. q. [s = 3@2 A 1].. S 1 1 1 ? 1, so the series q = q = [(q + 1) @q] lim (1 + 1@q) h q=1. . q. q q+1. q2. converges by the Root Test. 37. lim. q. s q S q q |dq | = lim (21@q 1) = 1 1 = 0 ? 1, so the series 2 1 converges by the Root Test. q. q=1. 11.8 Power Series 1. A power series is a series of the form. S. q=0. fq {q = f0 + f1 { + f2 {2 + f3 {3 + · · · , where { is a variable and the fq ’s are. constants called the coefcients of the series. S q 2 More generally, a series of the form q=0 fq ({ d) = f0 + f1 ({ d) + f2 ({ d) + · · · is called a power series in ({ d) or a power series centered at d or a power series about d, where d is a constant. q. . . . q+1. . . . dq+1 { { q |{| { = |{|. = lim = lim · 3. If dq = , then lim = lim s q dq q q + 1 {q q q + 1@ q q 1 + 1@q q By the Ratio Test, the series. {q S converges when |{| ? 1, so the radius of convergence U = 1. Now we’ll check the q q=1. endpoints, that is, { = ±1. When { = 1, the series the series. S 1 diverges because it is a s-series with s = q q=1. 1 2. 1. When { = 1,. (1)q S converges by the Alternating Series Test. Thus, the interval of convergence is L = [1> 1). q q=1. (1)q1 {q , then q3 & % 3 3 dq+1 (1)q {q+1 (1){q3 q q = lim = lim = lim · |{| = 13 · |{| = |{|. By the lim q dq q (q + 1)3 (1)q1 {q q (q + 1)3 q q+1. 5. If dq =.

(29) F. TX.10 Ratio Test, the series. POWER SERIES. ¤. 481. (1)q1 {q S converges when |{| ? 1, so the radius of convergence U = 1. Now we’ll check the q3 q=1. endpoints, that is, { = ±1. When { = 1, the series the series. SECTION 11.8. (1)q1 S converges by the Alternating Series Test. When { = 1, q3 q=1. (1)q1 (1)q 1 S S = converges because it is a constant multiple of a convergent s-series [s = 3 A 1]. 3 3 q q=1 q=1 q. Thus, the interval of convergence is L = [1> 1]. 7. If dq =. q+1 dq+1 { 1 q! {q = |{| lim = lim { , then lim · = |{| · 0 = 0 ? 1 for all real {. = lim q q q + 1 q! dq q (q + 1)! {q q q + 1 . So, by the Ratio Test, U = and L = (> ). q2 {q , then 2q % 2 & dq+1 (q + 1)2 {q+1 {(q + 1)2 |{| 1 2q |{| 2 lim 1+ (1) = = lim = lim · 2 q = lim = q dq q q 2q+1 q { 2q2 q 2 q 2. 9. If dq = (1)q. Ratio Test, the series. S. (1)q. q=1. When { = ±2, both series. S. q2 {q converges when 2q. (1)q. q=1. 1 2. 1 2. |{|. By the. |{| ? 1 |{| ? 2, so the radius of convergence is U = 2.. S q2 (±2)q = (

(30) 1)q q2 diverge by the Test for Divergence since 2q q=1. lim (

(31) 1)q q2 = . Thus, the interval of convergence is L = (2> 2).. q. u q+1 q+1 4 dq+1 q q (2)q {q = lim 2 |{| = 2 |{|, so by the Ratio Test, the · q q = lim 2 |{| 4 11. dq = , so lim 4 4 q q q d 2 |{| q + 1 q + 1 q q series converges when 2 |{| ? 1 |{| ? 12 , so U = 12 . When { = 12 , we get the divergent s-series s=. S q=1. 1 4 q. (1)q S 1 . When { = 12 , we get the series , which converges by the Alternating Series Test. 4 q q=1 Thus, L = 12 > 12 . 1 4. dq+1 {q+1 {q ln q 4q ln q |{| |{| , then lim · lim = ·1 = lim = 13. If dq = (1) q q 4 ln q dq q 4q+1 ln(q + 1) {q 4 q ln(q + 1) 4 q. [by l’Hospital’s Rule] = { = 4,. S. (1)q. q=2. |{| |{| . By the Ratio Test, the series converges when ?1 4 4. (1)q. q=2. |{| ? 4, so U = 4. When. [(1)(4)]q 1 S S S 1 1 {q 1 = = . Since ln q ? q for q 2, A and is the q ln q q=2 4 ln q ln q q q=2 ln q q=2 q. 4q. divergent harmonic series (without the q = 1 term), S. S q=2. 1 is divergent by the Comparison Test. When { = 4, ln q. S {q 1 = , which converges by the Alternating Series Test. Thus, L = (4> 4]. (1)q 4q ln q q=2 ln q.

(32) F. 482. ¤. TX.10. INFINITE SEQUENCES AND SERIES. CHAPTER 11. q+1 dq+1 q2 + 1 q2 + 1 ({ 2)q = lim ({ 2) , then lim · = |{ 2|. By the = |{ 2| lim 15. If dq = 2 2 q q q q q +1 dq (q + 1) + 1 ({ 2) (q + 1)2 + 1 ({ 2)q S converges when |{ 2| ? 1 [U = 1] 1 ? { 2 ? 1 1 ? { ? 3. When 2 q=0 q + 1. Ratio Test, the series { = 1, the series. S. (1)q. q=0. q2. comparison with the p-series. S 1 1 converges by the Alternating Series Test; when { = 3, the series converges by 2 +1 q=0 q + 1. 1 S [s = 2 A 1]. Thus, the interval of convergence is L = [1> 3]. 2 q=1 q. 3q+1 ({ + 4)q+1 dq+1 3q ({ + 4)q q q = lim , then lim = 3 |{ + 4| lim 17. If dq = · = 3 |{ + 4|. q q dq q 3q ({ + 4)q q+1 q+1 q By the Ratio Test, the series 13 ? { + 4 ?. 1 3. 3q ({ + 4)q S converges when 3 |{ + 4| ? 1 |{ + 4| ? q=1 q. 11 13 13 3 ? { ? 3 . When { = 3 , the series. , the series Test; when { = 11 3 19. If dq =. S 1 diverges s = q=1 q. 1 2. 1 3. U = 13. S. 1 (1)q converges by the Alternating Series q=1 q. . 1 . Thus, the interval of convergence is L = 13 > 11 3 3. s ({ 2)q |{ 2| = 0, so the series converges for all { (by the Root Test). , then lim q |dq | = lim q q qq q. U = and L = (> ). q ({ d)q , where e A 0. eq q+1 dq+1 1 |{ d| |{ d| eq = lim (q + 1) |{ d| 1 + = . · = lim lim q q dq q eq+1 q |{ d|q q e e. 21. dq =. By the Ratio Test, the series converges when. |{ d| ? 1 |{ d| ? e [so U = e] e ? { d ? e e. d e ? { ? d + e. When |{ d| = e, lim |dq | = lim q = , so the series diverges. Thus, L = (d e> d + e). q. q. dq+1 (q + 1)! (2{ 1)q+1 = lim (q + 1) |2{ 1| as q = lim 23. If dq = q! (2{ 1) , then lim q q dq q q!(2{ 1)q

(33) for all { 6= 12 . Since the series diverges for all { 6= 12 , U = 0 and L = 12 . q. dq+1 |4{ + 1|q+1 |4{ + 1| q2 = lim 25. lim · = |4{ + 1|, so by the Ratio Test, the series = lim q q (1 + 1@q)2 dq q |4{ + 1|q (q + 1)2 converges when |4{ + 1| ? 1 1 ? 4{ + 1 ? 1 2 ? 4{ ? 0 12 ? { ? 0, so U = 14 . When { = 12 , (1)q 1 S S , which converges by the Alternating Series Test. When { = 0, the series becomes , 2 2 q q=1 q=1 q a convergent s-series [s = 2 A 1]. L = 12 > 0 .. the series becomes.

(34) F. TX.10. SECTION 11.8. POWER SERIES. ¤. 483. q. { , then 1 · 3 · 5 · · · · · (2q 1) dq+1 {q+1 |{| 1 · 3 · 5 · · · · · (2q 1) = lim lim · = 0 ? 1. Thus, by the lim = q q dq q 1 · 3 · 5 · · · · · (2q 1)(2q + 1) {q 2q + 1. 27. If dq =. S. {q converges for all real { and we have U = and L = (> ). q=1 1 · 3 · 5 · · · · · (2q 1). Ratio Test, the series. S. q q=0 fq {. is convergent for { = 4. So by Theorem 3, it must converge for at least S q 4 ? { 4. In particular, it converges when { = 2; that is, q=0 fq (2) is convergent.. 29. (a) We are given that the power series. (b) It does not follow that. S. q q=0 fq (4). is necessarily convergent. [See the comments after Theorem 3 about convergence at. the endpoint of an interval. An example is fq = (1)q @(q4q ).] 31. If dq =. (q!)n q { , then (nq)! n dq+1 (q + 1)n = lim [(q + 1)!] (nq)! |{| = lim lim |{| n q q (q!) [n(q + 1)]! q (nq + n)(nq + n 1) · · · (nq + 2)(nq + 1) dq (q + 1) (q + 1) (q + 1) ··· |{| = lim q (nq + 1) (nq + 2) (nq + n) q+1 q+1 q+1 = lim lim · · · lim |{| q nq + 1 q nq + 2 q nq + n n 1 |{| ? 1 |{| ? nn for convergence, and the radius of convergence is U = nn = = n. 33. No. If a power series is centered at d, its interval of convergence is symmetric about d. If a power series has an innite radius. of convergence, then its interval of convergence must be (> ), not [0> ). (1)q {2q+1 , then q!(q + 1)! 22q+1 dq+1 {2q+3 1 q!(q + 1)! 22q+1 { 2 = lim lim = 0 for all {. · lim = 2 q q dq q (q + 1)!(q + 2)! 22q+3 {2q+1 (q + 1)(q + 2). 35. (a) If dq =. So M1 ({) converges for all { and its domain is (> ). (b), (c) The initial terms of M1 ({) up to q = 5 are d0 = d1 = . { , 2. {3 {5 {7 {9 , d2 = , d3 = , d4 = , 16 384 18,432 1,474,560. and d5 = . {11 . The partial sums seem to 176,947,200. approximate M1 ({) well near the origin, but as |{| increases, we need to take a large number of terms to get a good approximation..

(35) F. 484. ¤. INFINITE SEQUENCES AND SERIES. CHAPTER 11. 37. v2q1 = 1 + 2{ + {2 + 2{3 + {4 + 2{5 + · · · + {2q2 + 2{2q1. = 1(1 + 2{) + {2 (1 + 2{) + {4 (1 + 2{) + · · · + {2q2 (1 + 2{) = (1 + 2{)(1 + {2 + {4 + · · · + {2q2 ) 1 + 2{ 1 {2q [by (11.2.3)] with u = {2 ] as q [by (11.2.4)], when |{| ? 1. 1 {2 1 {2 1 + 2{ 1 + 2{ = v2q1 + {2q since {2q 0 for |{| ? 1. Therefore, vq since v2q and v2q1 both 1 {2 1 {2. = (1 + 2{) Also v2q. 1 + 2{ 1 + 2{ as q . Thus, the interval of convergence is (1> 1) and i ({) = . 1 {2 1 {2 s s S 39. We use the Root Test on the series fq {q . We need lim q |fq {q | = |{| lim q |fq | = f |{| ? 1 for convergence, or approach. q. q. |{| ? 1@f, so U = 1@f. S. fq {q diverges and. S. gq {q converges. By Exercise 11.2.69, S converge for |{| ? 2, the radius of convergence of (fq + gq ) {q is 2.. 41. For 2 ? { ? 3,. S (fq + gq ) {q diverges. Since both series. 11.9 Representations of Functions as Power Series 1. If i ({) =. S q=0. fq {q has radius of convergence 10, then i 0 ({) =. S q=1. qfq {q1 also has radius of convergence 10 by. Theorem 2. 1 , and then use Equation (1) to represent the function as a sum of a power 1u S S = ({)q = (1)q {q with |{| ? 1 |{| ? 1, so U = 1 and L = (1> 1).. 3. Our goal is to write the function in the form. series. i ({) =. 5. i ({) =. 1 1 = 1+{ 1 ({). 2 2 = 3{ 3. . 1 1 {@3. =. q=0. q=0. { { q S 1 2 S q or, equivalently, 2 { . The series converges when ? 1, q+1 3 q=0 3 3 q=0 3. that is, when |{| ? 3, so U = 3 and L = (3> 3). q S { {2q {2q+1 { 2 1 1 { S { { { S = = = = (1)q q = (1)q q+1 9 + {2 9 1 + ({@3)2 9 1 {({@3)2 } 9 q=0 3 9 q=0 9 9 q=0 q 2 { 2 S { 2 ? 1 { ? 1 |{|2 ? 9 |{| ? 3, so The geometric series converges when 3 3 9 q=0. 7. i ({) =. U = 3 and L = (3> 3). S S S S S S 1 1+{ = (1 + {) = (1 + {) 9. i ({) = {q = {q + {q+1 = 1 + {q + {q = 1 + 2 {q . 1{ 1{ q=0 q=0 q=0 q=1 q=1 q=1 The series converges when |{| ? 1, so U = 1 and L = (1> 1). A second approach: i ({) =. S S (1 {) + 2 1 1+{ = = 1 + 2 = 1 + 2 {q = 1 + 2 {q . 1{ 1{ 1{ q=0 q=1. A third approach:. 1 1+{ = (1 + {) i({) = = (1 + {)(1 + { + {2 + {3 + · · · ) 1{ 1{ = (1 + { + {2 + {3 + · · · ) + ({ + {2 + {3 + {4 + · · · ) = 1 + 2{ + 2{2 + 2{3 + · · · = 1 + 2. S q=1. {q ..

(36) F. REPRESENTATIONS OF FUNCTIONS AS POWER SERIES. SECTION 11.9. 11. i ({) =. 3 D E 3 = = + {2 { 2 ({ 2)({ + 1) {2 {+1. ¤. 485. 3 = D({ + 1) + E({ 2). Let { = 2 to get D = 1 and. { = 1 to get E = 1. Thus { q S 3 1 1 1 1 1 S 1 = = = ({)q 2 { {2 {2 {+1 2 1 ({@2) 1 ({) 2 q=0 2 q=0 q S S 1 1 1 q q q+1 = 1(1) { = q+1 {q (1) 2 2 2 q=0 q=0 We represented i as the sum of two geometric series; the rst converges for { (2> 2) and the second converges for (1> 1). Thus, the sum converges for { (1> 1) = L. 1 g S 1 g q q = = (1) { [from Exercise 3] g{ 1 + { g{ q=0 (1 + {)2 S S (1)q+1 q{q1 [from Theorem 2(i)] = (1)q (q + 1){q with U = 1. =. 13. (a) i ({) =. q=1. q=0. In the last step, note that we decreased the initial value of the summation variable q by 1, and then increased each occurrence of q in the term by 1 [also note that (1)q+2 = (1)q ]. 1 1 1 g 1 g S q q (b) i ({) = = (1) (q + 1){ = [from part (a)] 2 g{ (1 + {)2 2 g{ q=0 (1 + {)3 S S (1)q (q + 1)q{q1 = 12 (1)q (q + 2)(q + 1){q with U = 1. = 12 q=1. q=0. {2 1 1 S = {2 · = {2 · (1)q (q + 2)(q + 1){q 3 3 (1 + {) (1 + {) 2 q=0 1 S = (1)q (q + 2)(q + 1){q+2 2 q=0. (c) i ({) =. [from part (b)]. To write the power series with {q rather than {q+2 , we will decrease each occurrence of q in the term by 2 and increase the initial value of the summation variable by 2. This gives us ] 15. i ({) = ln(5 {) = . 1 g{ = 5{ 5. ]. 1 g{ = 1 {@5 5. 1 S (1)q (q)(q 1){q with U = 1. 2 q=2. { q S S 1 S {q+1 {q = F g{ = F q 5 q=0 5q (q + 1) q=0 5 q=1 q 5. ] . Putting { = 0, we get F = ln 5. The series converges for |{@5| ? 1 |{| ? 5, so U = 5. 17.. { { q S 1 1 q 1 1 S = = = { for ? 1 |{| ? 2. Now q+1 2{ 2(1 {@2) 2 q=0 2 2 q=0 2 q+1 S S q q1 1 1 g S 1 q g = = { { = {q . So 2 = q+1 q+1 q+2 g{ 2 { g{ q=0 2 ({ 2) q=1 2 q=0 2 i ({) =. q+1 q2 S S {3 3 S q+1 q { = {q+3 or {q for |{| ? 2. Thus, U = 2 and L = (2> 2). 2 = { q+2 q+2 q1 ({ 2) q=0 2 q=0 2 q=3 2.

(37) F. 486. ¤. CHAPTER 11. TX.10. INFINITE SEQUENCES AND SERIES. . 2 q S 1 1 { { S { S = = (1)q q {2q = (1)q q+1 {2q+1 . 16 q=0 16 16 q=0 16 16 q=0. 1 1 ({2 @16) { , The series converges when {2@16 ? 1 {2 ? 16 |{| ? 4, so U = 4. The partial sums are v1 = 16. 19. i ({) =. { { = {2 + 16 16. . v2 = v1 . {3 {5 {7 {9 , v = v + , v = v , v = v + , = = = . Note that v1 corresponds to the rst term of the innite 3 2 4 3 5 4 162 163 164 165. sum, regardless of the value of the summation variable and the value of the exponent.. As q increases, vq ({) approximates i better on the interval of convergence, which is (4> 4). ] ] ] ] 1+{ g{ g{ g{ g{ + = + = ln(1 + {) ln(1 {) = 1{ 1+{ 1{ 1 ({) 1{ ] ] S S = (1)q {q + {q g{ = [(1 { + {2 {3 + {4 · · · ) + (1 + { + {2 + {3 + {4 + · · · )] g{ . 21. i ({) = ln. ] =. q=0. q=0. (2 + 2{2 + 2{4 + · · · ) g{ =. ]. S q=0. 2{2q g{ = F +. But i (0) = ln 11 = 0, so F = 0 and we have i ({) =. 2{2q+1 S S 1 with U = 1. If { = ±1, then i ({) = ±2 , 2q + 1 2q +1 q=0 q=0. which both diverge by the Limit Comparison Test with eq = v3 = v2 +. 2{2q+1 S q=0 2q + 1. 1 2{ 2{3 . The partial sums are v1 = , v2 = v1 + , q 1 3. 2{5 , ===. 5. As q increases, vq ({) approximates i better on the interval of convergence, which is (1> 1)..

(38) F. TX 10 11.9 SECTION. REPRESENTATIONS OF FUNCTIONS AS POWER SERIES. ¤. 487. ]. 23.. w8q+2 S S S w 1 1 w . The series for =w· =w (w8 )q = w8q+1 gw = F + converges 8 8 8 1w 1w 1w 1 w8 q=0 q=0 q=0 8q + 2 when w8 ? 1 |w| ? 1, so U = 1 for that series and also the series for w@(1 w8 ). By Theorem 2, the series for ] w gw also has U = 1. 1 w8 S. {2q+1 with U = 1, so 2q + 1 q=0 S {2q+1 {5 {7 {3 {5 {7 {3 + + ··· = + ··· = and { tan1 { = { { (1)q+1 3 5 7 3 5 7 2q + 1 q=1. 25. By Example 7, tan1 { =. (1)q. [ {2q2 { tan1 { , so = (1)q+1 {3 2q +1 q=1. ]. 27.. S S {2q1 {2q1 { tan1 { =F+ . By Theorem 2, U = 1. g{ = F + (1)q+1 (1)q+1 2 3 { (2q + 1)(2q 1) 4q 1 q=1 q=1. q S S 1 1 = = (1)q {5q {5 = 5 5 1+{ 1 ({ ) q=0 q=0 ] ] S S 1 {5q+1 . Thus, g{ = (1)q {5q g{ = F + (1)q 5 1+{ 5q + 1 q=0 q=0 0=2 ] 0=2 {11 (0=2)11 {6 (0=2)6 1 + · · · + · · · . The series is alternating, so if we use L= g{ = { = 0=2 5 1 + { 6 11 6 11 0 0. the rst two terms, the error is at most (0=2)11@11 1=9 × 109 . So L 0=2 (0=2)6@6 0=199989 to six decimal places. 29. We substitute 3{ for { in Example 7, and nd that. ]. ]. { arctan(3{) g{ =. { ]. So 0. S. (1)q. q=0. 0=1. (3{)2q+1 g{ = 2q + 1 . { arctan(3{) g{ = =. ]. S. (1)q. q=0. S 32q+1 {2q+2 32q+1 {2q+3 g{ = F + (1)q 2q + 1 (2q + 1)(2q + 3) q=0. 3{3 33 {5 35 {7 37 {9 + +··· 1·3 3·5 5·7 7·9. 0. 0=1. { arctan(3{) g{ . 0. 1 9 243 2187 + +···. 103 5 × 105 35 × 107 63 × 109. The series is alternating, so if we use three terms, the error is at most ]. 0=1. 2187 3=5 × 108 . So 63 × 109. 1 9 243 + 0=000 983 to six decimal places. 103 5 × 105 35 × 107. 31. Using the result of Example 6, ln(1 {) = . ln 1=1 = ln[1 (0=1)] = 0=1 . {q S , with { = 0=1, we have q=1 q. 0=001 0=0001 0=00001 0=01 + + · · · . The series is alternating, so if we use only 2 3 4 5. the rst four terms, the error is at most. 0=01 0=001 0=0001 0=00001 = 0=000002. So ln 1=1 0=1 + 0=09531. 5 2 3 4.

(39) F. 488. ¤. INFINITE SEQUENCES AND SERIES. CHAPTER 11. 33. (a) M0 ({) =. TX.10. (1)q {2q (1)q 2q{2q1 (1)q 2q(2q 1){2q2 S S S > M00 ({) = , and M000 ({) = , so 2q 2 2q 2 2 (q!) 22q (q!)2 q=0 2 (q!) q=1 q=1. {2 M000 ({) + {M00 ({) + {2 M0 ({) = =. (1)q 2q(2q 1){2q (1)q 2q{2q (1)q {2q+2 S S S + + 2q 2 2q 2 2 (q!) 2 (q!) 22q (q!)2 q=1 q=1 q=0 (1)q 2q(2q 1){2q (1)q 2q{2q S S S (1)q1 {2q + + 2q 2 2q 2 2q2 2 (q!) 2 (q!) [(q 1)!]2 q=1 q=1 q=1 2. (1)q 2q(2q 1){2q (1)q 2q{2q (1)q (1)1 22 q2 {2q S S S + + 22q (q!)2 22q (q!)2 22q (q!)2 q=1 q=1 q=1 S 2q(2q 1) + 2q 22 q2 2q = (1)q { 22q (q!)2 q=1 2 S 4q 2q + 2q 4q2 2q { =0 = (1)q 22q (q!)2 q=1. =. ]. 1. (b) 0. ] M0 ({) g{ =. 1. 0. . ] 1 (1)q {2q S {4 {6 {2 + + · · · g{ g{ = 1 2q 2 4 64 2304 q=0 2 (q!) 0. 1 {3 {5 {7 1 1 1 = { + +··· = 1 + + ··· 3·4 5 · 64 7 · 2304 12 320 16,128 0 Since U1 0. 1 16,128. 0=000062, it follows from The Alternating Series Estimation Theorem that, correct to three decimal places,. M0 ({) g{ 1 . 35. (a) i ({) =. {q S q=0 q!. 1 12. 1 320. +. 0=920.. i 0 ({) =. q{q1 {q S S S {q1 = = = i({) q! q=1 q=1 (q 1)! q=0 q!. (b) By Theorem 9.4.2, the only solution to the differential equation gi ({)@g{ = i ({) is i ({) = Nh{ , but i (0) = 1, so N = 1 and i ({) = h{ . Or: We could solve the equation gi ({)@g{ = i ({) as a separable differential equation. q+1 2 dq+1 q {q q2 = lim { = |{| lim , then by the Ratio Test, lim · = |{| ? 1 for q dq q (q + 1)2 q q + 1 q2 {q {q S = S 1 which is a convergent s-series (s = 2 A 1), so the interval of convergence, so U = 1. When { = ±1, q2 2 q=1 q=1 q. 37. If dq =. convergence for i is [1> 1]. By Theorem 2, the radii of convergence of i 0 and i 00 are both 1, so we need only check the endpoints. i ({) =. {q S 2 q=1 q. i 0 ({) =. q{q1 S S {q , and this series diverges for { = 1 (harmonic series) = q2 q=1 q=0 q + 1. and converges for { = 1 (Alternating Series Test), so the interval of convergence is [1> 1). i 00 ({) = at both 1 and 1 (Test for Divergence) since lim. q. q{q1 S diverges q=1 q + 1. q = 1 6= 0, so its interval of convergence is (1> 1). q+1.