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# SEQ, SERIES, AND MULTIVAR CALC M 408D

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This 20 page Class Notes was uploaded by Reyes Glover on Sunday September 6, 2015. The Class Notes belongs to M 408D at University of Texas at Austin taught by John Gilbert in Fall. Since its upload, it has received 36 views. For similar materials see /class/181428/m-408d-university-of-texas-at-austin in Mathematics (M) at University of Texas at Austin.

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Date Created: 09/06/15

moseley cmm3869 7 HW02 7 Gilbert 7 56380 This print out should have 18 questions Multiple choice questions may continue on the next column or page 7 find all choices before answering 001 100 points Use double angle formulas to simplify the expression e 3cose7sine2 34sin2e5cos2e 1 M 2 e 735sin2e4cos2e 3 g 4 e 5 7 3 sin 2e 1 4 cos 2e correct 473sin2e5cos2e 5 f e 5 4sin2e 7 3cos2e 6 e 4 4sin2e 7 3cos2e Explanation After expansion we see that e 9 cos2 e 7 6sine cose 1 sin2 e But cos2e 1 cos 2e 1 2 1 7 cos 2e 0M7 sin2 e while 2sine cose sin2e Consequently e 573sin2e4cos2e 002 100 points When is the graph of y abcosmx m gt 0 on 74 4 what is b 1 b 3 correct 5 2 b Z 3 b 73 4 b 77 5 b 3 Explanation As shows the given graph is that of 71 cos 7rx y 2 2 a in other words the graph of 3 3 COS W y 2 2 shifted vertically by a term y 71 Thus b is given by 003 part 1 of 3 100 points The figure AB C D in moseley cmm3869 7 HW02 7 Gilbert 7 56380 2 O is a parallelogram and ED is a diagonal i Express the area of ABC39D as a function of a b and 6 1 Area1301 abcos l 1 2 AreaABCD iabsin l l 3 AreaABCD Eabcos l 4 Area1301 absin l correct 5 Area1301 2absin6l Explanation In the parallelogram ABC39D the triangles AABD and ABDC are congruent by SSS since side ED is common to both triangles while side AD is congruent to side BC and side AB is congruent to side DO Thus Area1301 Area131 AreaBDc 2AreaABD Now let h be the height of the perpendicular from D onto side AB of as shown in grey in Then h asin l is the height of AABD while I is its base length Thus 1 l AreaABD ibase gtltheight Eabsin Consequently Area1301 absin l 004 part 2 of 3 100 points ii Find 6 when a 6 b 8 and the length of the diagonal ED is 7 1 6 60910 2 6 59910 3 6 58910 4 6 5791O correct 5 9 6191O Explanation To determine 6 we use the Law of Cosines with c the length of diagonal BD 02 a2 b2 7 2abcos6 When therefore 36 64 7 49 7 cos 6 96 Inoseley cnnn3869 7 HW02 7 Gilbert 7 56380 3 9 i 17 i 57 910 i arccos 32 i 005 part 3 of 3 100 points iii Then find the area ofparallelograrn ABOD 1 AreaABCD 4092 sq units 2 Area1301 4117 sq units 3 AreaABCD 4142 sq units 4 AreaABCD 4067 sq units correct 5 Area1301 4167 sq units Explanation By parts and ii AreaABCD ab sin l 48 sin5791o Consequently Area1301 4067 sq units 006 100 points Find the derivative of f When m 7 2 1 fac 47281111620 3 far lisifl correct 4 m 7 5 WW sinx72 cos2 x 27cosac 5 fQC f sin at 2sinac 7 7 f cos2 x 1 2 cosx 8 fQC if sin at Explanation By the quotient rule sin2 x 7 cos x2 7 cos x sin2 x fyc i singx cosgac 7 2cosac sin2 x But cos2 x 717 sin2 x 1 Consequently 1 7 2cosac f W 2 sin x 007 100 points A particle moves along a straight line so that its acceleration at any given time t is at 2t 7 12 in units of feet and seconds If the initial velocity of the particle is 32 feetsec at What time If during the interval 0 g t g 7 does the particle change direction 1 t 3seconds 2 t 1seconds 3 t 4 seconds correct 4 none of the other answers 5 t 5 seconds 6 t 2seconds Explanation The acceleration at time t is given by at g 22412 moseley cmm3869 7 HW02 7 Gilbert 7 56380 4 Thus was t2 i 1225 0 Where O is determined by the initial condition 110 32 In this case 0 32 so vt t2 712t32 Now the particle Will change direction at a critical point of its position function 5t 26 at a solution of ds 157 75741578 0 vlt gt d lt gtlt gt Consequently the particle Will be furthest to the right during the interval 0 g t g 7 When keywords acceleration velocity antideriva tive 008 100 points A cap of a sphere is generated by rotating the shaded region in about the y axis Determine the volume of this cap When the radius of the sphere is 6 inches and the height of the cap is 1 inch 16 1 volume En cu ins 2 volume g7 cu ins correct 3 volume 57139 cu ins 4 volume 67139 cu ins 5 volume 1337 cu ins Explanation Since the sphere has radius 7 6 inches we can think of this sphere as being generated by rotating the circle 2 y2 62 about the y aXis The cap of the sphere is then generated by rotating the the graph of x y V627y2 on the interval 5 6 about the y axis Thus the volume of the cap is 6 V n5 fwdy 6 1 6 7r 627y2dy 7r 622477243 5 3 5 Consequently 17 3 keywords volume integral solid of revolu tion integral respect to y 009 100 points The volume of a solid can often be deter mined by integration even When the cross section of the solid is not necessarily a circle In the following figure l J nuanwamwx 1 lllhil l l h moseley cmm3869 7 HW02 7 Gilbert 7 56380 the base is the circle 2y2 4 and the cross section perpendicular to the y aXis is a square Set up a de nite integral expressing the volume V of the solid 2 1 volume 44y2dy cuunits 72 2 2 volume 4y2dy cuunits 72 2 3 volume 47y2dy cuunits 72 2 4 volume 447y2 dy cuunits 2 correct 2 5 volume 247y2dy cuunits 72 Explanation At the point P at y on the base of the solid the cross section is a square having side length 215 Thus the area of cross section is a square having area 433 so the volume of the solid is given by the definite integral 2 V 42dy 72 Hence 2 volume 447y2dy cuunits 72 keywords volume integral cross section 010 100 points Which of the following is the graph of ne 3x73 moseley cmm3869 7 HW02 7 Gilbert 7 56380 6 correct Explanation Since lim 3 0 7700 we see that lim at 73 7700 in particular f has a horizontal asymptote y 73 This eliminates all but two of the graphs On the other hand f0 72 so the y intercept of the given graph must occur at y 72 Consequently the graph is of f is 011 100 points Determine if 367 46 hm x7gtoo 56 7 67 exists and if it does find its value 3 1 limit 7 5 2 limit 0 4 3 hmit 5 correct 4 limit does not exist 5 limit 73 6 limit 74 Explanation Since lim 6quot 0 lim 6 oo IHOO I lOO when a gt 0 taking limits directly gives 367 46 oo hm 7 7 x700 56 7 6 x 00 which doesn7t make any sense And we can7t just cancel the 007s because infinities don7t work like that So we try to get rid of terms that go to 00 and leave terms that go to zero To achieve this multiply top and bottom by 6 Then 6 367 46 i 3672i 4 6 7 567 7 6 7 7 5 7 6 27 moseley cmm3869 7 HW02 7 Gilbert 7 56380 In this case 36 717 46 H1 x700 56 7 67 35 4 700 5 7672 But as we noted lim 6 29 0 I 00 so by properties of limits lim 36 4 4 I 00 lim 5 4 5 29 5 IHOO Thus by properties of limits yet again 367 46 lim x700 55 7 exists and the 4 5 keywords exponential function limit at in finity properties of limits 012 100 points Find the derivative of fac 2 310 167I 1 far 4 7 an 7 267I correct 2 far 1 7 x 7 267I 3 far 4 7 310 7 267I 4 far 4 7 310 267I 5 far 1 7 310 267I 6 far 1 7 x 267I Explanation By the Product Rule far 23E 367I 7102 320 7 1e Consequently 013 100 points Find When dac ylnac7y2 33c 1 dy 37y die 7 lnx72y 2 n 3y dac lnx72y 3 dy i 37y 39 die 7 lnx2acy 4 3gE iycorrect dac arlnac72xy 5 dy 7 3w die 7 aclnac2y dy 3y 39 g i atlnab 717 2x1 Explanation Differentiating implicitly With repect x we see that dy dy yltgtlnac72yE 3 Which after rearrangement becomes dy y l 7 2 7 3 7 7 n96 yd 6 Consequently dy 7 3x72 die 7 ln72xy 014 100 points Determine the integral 1 I die o 47102 Inoseley cnnn3869 7 HW02 7 Gilbert 7 56380 2 1 I 7 3 1 2 I 7 3 3 1 l 2 4 I 17139 2 1 5 I g7 correct 2 6 I g7 Explanation Since 1 d 39 1 0 7 ac sin ac V17202 we need to reduce I to an integal of this form by changing the at variable Indeed set x 21L Then dac 2 du while In this case 12 1 I 4 du o 2V1 7 u 12 1 2 idu o x17u2 Consequently keywords 015 100 points Evaluate the integral 7r2 I singacdx O 1 1 l 6 2 1 6 1 3 I 7 3 2 4 I 7 correct 3 5 I 1 Explanation Since singx 17cos2x we see that 7r2 I 17cos2sindac 0 This suggests using the substitution u cos at For then du 7 sinac die while x0 gt u1 x7 gt In this case 016 100 points Evaluate the integral 1 7r3 I 1sin6d6 0 1 1 I 371 21 2 moseley cmm3869 7 HW02 7 Gilbert 7 56380 9 l 3 I EC 7 correct 4 I 1 5 I 2 6 I 1 Explanation By the double angle formula sin2ac 2sinac cosx h39l b th P th 39d t39t W 18 y e y agoreanl en 1 y Determine the largest possible volume the COS2 35 Sin2 an 1 cylinder can have 288 Thus 1 max volume 7 7r cu cms correct 1 sin 210 cos2 2 sinac cos at sin2 at 2 max volume 288 7r cu cms cos at sin an 72 Taking an 62 we thus see that 3 max volume E 7r cu cms 2 1 T 5mg Kw2 T Slng2 4 max volume Lg 7r cu cms cos62 sin62 5 max volume 144 7r cu cms whenever cos62 sin62 2 0 6 max volume 72 7r cu cms hence for 0 S 6 S 7139 Thus Explanation Wg Let the cylinder have radius 7 and height I cos 9 Sin d6 2h Then the relation between the dimensions 2 0 2 2 of the cylinder and sphere are shown in 6 6 7r3 sin 7 7 cos 7 2 2 0 Consequently I lt37 gt 017 100 points A right circular cylinder is inscribed in a sphere of diameter 12 cms as shown in moseley cmm3869 7 HW02 7 Gilbert 7 56380 10 Thus its volume is given by 2 2 2 I xgcorrect V0 h 71397quot gtlt 2h 27m It On the other hand by Pythagoras 339 I 1 r2h236 41 Thus as a function of 7 5 I 22 V0 27rrgx 6 I 2M6 and so Eg ngxigile angle formula for cos 220 Vr 27r27 736r 3j cos2x 2cos27l 2m 72 7 3702 so taking at 62 we see that V36 7 7 But then the only critical point that makes geometric sense is 7 6x2xg furthermore 9 9 1cos6 2cos2 2cos When cos62 2 0 hence for 0 S 6 S 7139 6 2 Thus 1 gt Vr gt 0 g27r3 xg 27r3 g I cos7d6l 22sin7 0 2 2 0 6 Consequently the maximum value of the vol ume occurs at 7quot 6x2xg and rlt While Consequently rgt gt Vrlt0 keywords de nite integral double angle for 288 max1mum volume 77139 cu cms xg mula keywords optimization maximum volume 3 space geometry 018 100 points Evaluate the integral 27r3 I 1cos6d6 0 Hint use a double angle formula to express 1 cos 6 in terms ofcos2 62 11 moseley cmm3869 7 HW01 7 Gilbert 7 56380 1 This print out should have 20 questions Multiple choice questions may continue on the next column or page 7 find all choices before answering Welcome to Quest Print o 39 this as signment and bring it to lectures as well as discussion groups 001 100 points Determine if 4x2 3 lim lt gt 771 2 1 exists and if it does find its value 7 1 limit E correct 2 limit 3 3 limit does not exist 4 limit 7 5 limit 4 Explanation Set at 4x23 910 2471 Then lim at 7 lim 910 2 771 771 Thus the limits for both the numerator and denominator exist and neither is zero so L7Hospital7s rule does not apply In fact all we have to do is use properties of limits For then we see that 7 lquott7 002 100 points Determine the value of lim 7 700 1 1 limit 1 correct 2 limit 0 3 limit l 4 4 limit 4 5 limit oo 6 limit 2 7 limit l 2 Explanation Since 1 at 00 im 7 7 i x700 m 00 the limit is of indeterminate form We might first try to use L7Hospital7s Rule we hm we 139 7 xgmoo gltgt z7gtoo gm with f MV 5 to evaluate the limit But 7 7 3 flt717 g 267 so 2 lim f lim in 6 00 x700 g 700 m 00 which is again of indeterminate form Let7s try using L7Hospital7s Rule again but now with 2476 ga and 7 1 26a 9H lnthiscase 2 lim lim I700 moseley cmm3869 7 HWOl 7 Gilbert 7 56380 2 Which is the limit we started With So this is Consequently an example Where L7Hospital7s Rule applies but doesn7t work 4 limit 7 We have to go back to algebraic methods 3 x at l 7 004 100 pomts xx26 lath16ac2 16ac2 for 95 gt 039 Thus When f g F and G are functions such that x l 139 7 139 xgo 26 igloo 16 2 1311 0 Illinl g 0 lim 2 lim C15 oo and so I 71 mai Which if any of 003 100 points A 11311 WWW Determine if the limit B hm 30996 1 7gt 1 4 x 7 l 39 7 996 9611311 36371 0 exists and if it does find its value are indeterminate forms 3 1 limit Z 1 A only 4 2 limit g correct 2 an Of them 3 limit oo 3 Bonly correct 4 none of the other answers 4 none of them 5 limit 700 Explanation 5 A and C only Set f41 g371 6 Conly Then 7 A and B only lim at 0 lim 910 0 I l I l 8 BandConly so L7Hospital7s rule applies Thus Explanation im im A By properties of limits H gltxgt H me But 0390 07 far 4mg gx 33c2 so this limit is not an indeterminate form moseley cmm3869 7 HW01 7 Gilbert 7 56380 3 B Since lim fxg 00 x7 1 this limit is an indeterminate form C By properties of limits 996 7 0 717 a i 0 so this limit is not an indeterminate form 005 100 points Find the value of 2x 7 lnac lim x7gtO 3 1 limit 2 2 2 limit 7 3 3 limit 7oo 4 limit oocorrect 5 none of the other answers 6 limit 0 7 limit 3 Explanation Let7s first check if the given limit is an indeterminate form Now hiac is defined for x gt 0 and the graph of lnac has a vertical asymptote at x 0 in addition lim lnac 7oo x7gt04r On the other hand 27lnx 2 lnx 3x Thus hm 27lnx 2g x7gtO 3x 3 0 which is not an indeterminate form In fact since the second term is positive we see that 210 7 ln x lim 957 0 320 006 100 points Determine if sin 13ac m I 7 0 tan 14ac exists and if it does nd its value 1 limit 3 4 2 limit 7 3 3 limit does not exist 3 4 limit 1 correct 5 limit 4 6 limit 0 Explanation Since the limit has the form 00 it is in determinate and L7Hospital7s Rule can be ap plied 1 06 f 1210 gx 7 1210 gx with we 7 sin1lt3xgt gltxgt 7 tan1lt4xgt and we 7 3 we 4 In this case we lt116m2gt m gx 7 4 179352 4 as x 7 0 Consequently the limit exists and 3 4 007 100 points moseley cmm3869 7 HWOl 7 Gilbert 7 56380 4 Determine hm x 3 7 4lnx x7gtO 1 limit 0c0rrect 2 limit oo 3 none of the other answers 4 limit 74 5 limit 71 6 limit 7oo Explanation Since the limit is of the form 0 00 use of L7Hospital7s Rule is suggested First write 374l x374lnx f1 E and set 1 at 374lnac 920 7 an Then H6 00 lin Jr g 005 so L7Hospital7s Rule applies lim fa hm f 7gtO 9a 9040 gx 4 lim 1 x7gtO 1 372 1 95 531 g Consequently hm x374lnx 0 x7gtO 008 100 points Determine if 5 5 lim 7 7 7 70 67071 exists and if it does nd its value 5 1 limit 7 4 2 hmit does not exist 3 l39 39t i 5 imi 7 3 4 limit 5 5 limit 0 5 6 limit E correct Explanation Now 5 5 6 7 l 7 x at 5lt gt x 6 7 1 Me 71 910 where f g are everywhere differentiable func tions such that lim at 0 lim 910 0 70 70 Thus L7Hospital7s rule can be applied But far 56 7 5 while 996 6I16I so K76 f BEIGE 7 1310 gx i 7 hm x70 6 are 7 1 As f and g are differentiable functions such that hm rm 7 0 x70 hm gx 0 x70 moseley cmm3869 7 HW01 7 Gilbert 7 56380 we have to apply L7Hospital7s rule again But We 7 56 from which it follows that g x 26 6I lim f x 5 lim g 2 70 70 Consequently the limit exists and 5 lquott7 009 100 points Determine if the limit lim 1 7mm 1 7gt 0 exists and if it does nd its value 1 none of the other answers 2 limit e 3 limit 7 4 limit e 5 limit 67 correct 6 limit oo 7 limit 7oo Explanation Because of the variable in the exponent we take logs and then take the limit and nally exponentiate to recover the original limit For then hm01 7CSCI elimxao lnl7x I7 But by properties of logs lnl 7mm csc lnl 79c which in turn can be written as f 900 where at lnl7ac 910 sinac Now lim lnl7x 0 x70 while lim sinac 0 Thus L7Hospital7s rule can be applied 1 06 WW 3310 m 3310 m 7 lim x70 17cosx Consequently the limit exists and 010 100 points Determine if the limit lim 1 7 830 I 700 exists and if it does nd its value 78 1 limit e 2 none of the other answers V we 3 limit 4 limit oo 5 limit 7oo 6 limit 0 7 limit lcorrect Explanation Since the limit has the form lim 17810 000 I H 700 we rst take logs and evaluate 90th ln 1 7 830 moseley cmm3869 7 HW01 7 Gilbert 7 56380 6 ln17 ln1 7 8x 510 By L7Hospital7s Rule therefore 90th ln17 8x hm ltgt 0 7700 5 178 Consequently the limit exists and limit 60 1 Which of the following integrals are im 011 100 poin proper H 9 1 9 3 9quot 100 1 1 1 96 2 72 12 9 d o il ts dab 7 1 1 I idm 3 0 V1 I3 only I2 and I3 only all of them I1 and I2 only I1 only I 2 only correct I1 and I3 only none of them Explanation An integral I ab facdx is improper When one or more of the following conditions are satis ed i the interval of integration is infinite 26 When a 700 or b 00 or both ii f has a vertical asymptote at one or moreofxaxboraccforsome a lt c lt b Consequently 100 1 7 die 1 1 2 is not improper 2 7 x 2th 0 71 is improper is not improper 012 100 points Determine if the improper integral 0 5 I 4 70543 dac converges and if it does compute its value 6 1 I 7 5 1 2 I 77 2 3 I doesn7t converge 1 4 I i correct moseley cmm3869 7 HW01 7 Gilbert 7 56380 7 5 5 I 7 ll Explanation The integral is improper because of the infi nite interval of integration To overcome this we set 1 I lim 7amp6 1700 4 62 Whenever the limit on the right hand side exists NOW Since lim 7 t7gtoo it follows that the improper integral converges and that 00 I del 4 956 2 013 100 points Find the total area under the graph of i 8 form 2 l 8 1 Area g correct 2 Area 00 7 3 Area 7 3 4 Area 3 5 Area 2 5 6 Area 7 3 Explanation The total area under the graph for x 2 l is an improper integral Whose value is the limit it 8 lim 74dac t7gtoo 1 t 8 t 874d 77x 3 1 3 1 Consequently 8 8 Area lim 7l7t73 7 1700 3 3 014 100 points Determine if 00 I 37 3 273 converges and if it does compute its value 623 1 I 7 4 23 2 I 73 6 4 3623 3 I 4 4 I 623 3623 5 I 77 2 6 I does not converge correct Explanation The integral I is improper because of the infinite interval of integration Thus I Will converge if t m thrglo 3 132 7 3 d moseley cmm3869 7 HW01 7 Gilbert 7 56380 8 exists To evaluate this last integral we use substitution setting it x 7 3 For then du2dx While x3 gt u6 xt gt 71715273 In this case 7 73 1 1 BLdQE 7 du 3 273 2 6 u u23t273 EU 7 323 7 623 6 However lim 5 7 323 oo t7gtoo Consequently I does not converge 015 100 points Determine if the improper integral 00 6t 71 I an fan 1 converges and if it does find its value 9 1 I 7772 8 3 2 I 7772 4 9 3 I E79 correct 3 4 I 7 4 5 I does not converge 9 6I7 16 Explanation The integral I is improper because the in terval of integration is in nite Thus the integral Will converge if lim It It t 7gt oo 1 exists and its value Will then be the value of the limit The substitution u tan 1 x is suggested For then t 6tan 1ac 12 dab du gdac 1 While and x t gt u tanilt In this case wilt tani1 t It 6udu 3u2 7r4 7r4 2 7 2 7r It 3tan1t On the other hand lim tanilt 5 t7gtoo 2 Consequently I converges and has value 016 100 points Determine if the improper integral 1 71 8sin x I 7 die 0 x 1 7 2 is convergent or divergent and if convergent nd its value 1 LI 2I1 moseley cmm3869 7 HWOl 7 Gilbert 7 56380 9 1 3 I 7W2 4 4 I is divergent 5 I 71392 correct Explanation The integral is improper because u sin 1 x 7 731117 11 7 2 7 Thus we have to check if t 71 8 sin x lim 7 die t7 17 0 11 7 2 is convergent or divergent To determine 8sin 1 v1 7 2 set it sin 1 x Then die 1 du idx V1720 and so 8sin71 2 Wdat i Sudu 7 4n 0 from which it follows that t 8 sin 1 an 7 die 0 x l 7 2 t 4 sin 1 3010 4sin 1t2 139 39 422 7 111 W gt Consequently I is convergent and 017 100 points Determine if the improper integral 1 3 ln 710 at 0 is convergent or divergent and if convergent nd its value 1 I 7ln77l 2 I 6ln772 correct 3 I 31n71 4 I 71n71 5 I 6ln72 6 I is divergent 7 I 3111772 Explanation The integral is improper because the inte grand has a vertical asymptote at x 0 and so is not bounded on 0 1 Thus 1 317 I lim n dx t70 t V5 TO evaluate the integral 1 3ln 710 It d 1 V5 we use Integration by Parts For then It 6 ln7x7t1 d 1 6Eln7ac77t dan 6Eln772E Thus It 6ln77 2 76xtln7t 72 moseley cmm3869 7 HW01 7 Gilbert 7 56380 10 But by L7Hospital7s Rule lim xEin 7t 0 t7gt0 Consequently I 6ln772 018 100 points Use the Comparison Theorem to determine Which of the following integrals is convergent 00 cos2 710 1 5 2 die correct 1 2 x 60 5 d 0 2 2 00 V1V3ac 3 7 3dac 1 at 00 4 5 d3 60 95 00 die 5 774116 1 V310 Explanation Equot cos2 710 1 F0 21 242 m S OO 1 72dac is convergent by Equation 2 With an 1 00 cos2 710 p 2 gt 1 so 5 2 die is convergent by 2 x the Comparison Theorem 019 100 points Determine if the integral 1 713 4 3 1 11 356 0 962 converges and if it does find its value 1I30 2I29 3 I does not converge 4 I 28 5 I 27 6 I 26 correct Explanation In the given integral 4 33513 4 3 912 mtm for x gt 0 so it is enough to check if the integral 1 1 II 7dx 0 95 converges for p lt 1 Now II will converge When the limit 1 1 lim 7dx t70 t 3519 exists But 11d 7 7p1 1 t E T7p1t 1 1111lt1 11gt 17p tIH lfplt1thenp71lt0so 1 lim 7 lim 251 0 t7gt0 tpil t7gt0 Thus Ip converges for all p lt 1 and 1 1 p 351 71Tt1 19 171 Since both exponents satisfy the condition 1 lt 1 I converges and 4 3 17 726 12 16 020 100 points

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