a 1 1 a 2 1 a a n 1 ò a n or o a n n n 1 32, 63 16, 31

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1 206 Cegage Learig. All Rights Reserved. This cotet is ot yet fial ad Cegage Learig Series The curret record for computig a decimal approximatio for was obtaied by Shigeru Kodo ad Alexader Yee i 20 ad cotais more tha 0 trillio decimal places. Sum of first terms What do we mea whe we express a umber as a ifiite decimal? For istace, what does it mea to write The covetio behid our decimal otatio is that ay umber ca be writte as a ifiite sum. Here it meas that where the three dots s d idicate that the sum cotiues forever, ad the more terms we add, the closer we get to the actual value of. ` I geeral, if we try to add the terms of a ifiite sequece ha j we get a expressio of the form a a 2 a a which is called a ifiite series (or just a series) ad is deoted, for short, by the symbol ò a or o a Does it make sese to talk about the sum of ifiitely may terms? It would be impossible to fid a fiite sum for the series because if we start addig the terms we get the cumulative sums,, 6, 0, 5, 2,... ad, after the th term, we get s dy2, which becomes very large as icreases. However, if we start to add the terms of the series we get 2, 4, 7 8, 5 6, 2, 6 64,..., 2 y2,.... The table shows that as we add more ad more terms, these partial sums become closer ad closer to. I fact, by addig sufficietly may terms of the series we ca make the partial sums as close as we like to. So it seems reasoable to say that the sum of this ifiite series is ad to write ò We use a similar idea to determie whether or ot a geeral series () has a sum. We cosider the partial sums s a s 2 a a 2 s a a 2 a ad, i geeral, s 4 a a 2 a a 4 s a a 2 a a o a i i

2 2 SERIES These partial sums form a ew sequece hs j, which may or may ot have a limit. If lim l ` s s exists (as a fiite umber), the, as i the precedig example, we call it the sum of the ifiite series o a. Compare with the improper itegral y` f sxd dx lim t l ` yt f sxd dx To fid this itegral we itegrate from to t ad the let t l `. For a series, we sum from to ad the let l `. 2 Defiitio Give a series o ǹ a a a 2 a, let s deote its th partial sum: s o a i a a 2 a i If the sequece hs j is coverget ad lim l ` s s exists as a real umber, the the series o a is called coverget ad we write a a 2 a s or ò a s The umber s is called the sum of the series. If the sequece hs j is diverget, the the series is called diverget. Thus the sum of a series is the limit of the sequece of partial sums. So whe we write o ǹ a s, we mea that by addig sufficietly may terms of the series we ca get as close as we like to the umber s. Notice that ò a lim l ` o a i i EXAMPLE Suppose we kow that the sum of the first terms of the series oǹ a is s a a 2 a The the sum of the series is the limit of the sequece hs j: ò a lim l ` s lim l ` 2 5 lim l ` I Example we were give a expressio for the sum of the first terms, but it s usually ot easy to fid such a expressio. I Example 2, however, we look at a famous series for which we ca fid a explicit formula for s. See also Sectio EXAMPLE 2 A importat example of a ifiite series is the geometric series a ar ar 2 ar ar 2 ò ar 2 a ± 0 Each term is obtaied from the precedig oe by multiplyig it by the commo ratio r. (We have already cosidered the special case where a 2 ad r 2 o page.) If r, the s a a a a l 6`. Sice lim l ` s does t exist, the geometric series diverges i this case. If r ±, we have s a ar ar 2 ar 2 ad rs ar ar 2 ar 2 ar 206 Cegage Learig. All Rights Reserved. This cotet is ot yet fial ad Cegage Learig

3 SERIES Figure provides a geometric demostratio of the result i Example 2. If the triagles are costructed as show ad s is the sum of the series, the, by similar triagles, s a a a 2 ar so s a 2 r Subtractig these equatios, we get s 2 rs a 2 ar s as 2 r d 2 r ar# If 2, r,, the r l 0 as l `, so 206 Cegage Learig. All Rights Reserved. This cotet is ot yet fial ad Cegage Learig a a-ar FIGURE a ar ar@ ar@ ar I words: The sum of a coverget geometric series is first term 2 commo ratio What do we really mea whe we say that the sum of the series i Example is? Of course, we ca t literally add a ifiite umber of terms, oe by oe. But, accordig to Defi i tio 2, the total sum is the limit of the sequece of partial sums. So, by takig the sum of sufficietly may terms, we ca get as close as we like to the umber. The table shows the first te partial sums s ad the graph i Figure 2 shows how the sequece of partial sums approaches. a s Thus whe r as 2 r d lim l ` s lim l ` 2 r a 2 r 2 a 2 r lim r l ` a 2 r, the geometric series is coverget ad its sum is ays 2 rd. If r < 2 or r., the sequece hr j is diverget ad so, by Equatio, lim l ` s does ot exist. Therefore the geometric series diverges i those cases. We summarize the results of Example 2 as follows. 4 The geometric series is coverget if r If r ò ar 2 a ar ar 2, ad its sum is ò ar 2 >, the geometric series is diverget. EXAMPLE Fid the sum of the geometric series a 2 r r, SOLUTION The first term is a 5 ad the commo ratio is r 2 2. Sice r 2,, the series is coverget by (4) ad its sum is s (2 2 ) 5 5 s 0 20 FIGURE 2

4 4 SERIES Aother way to idetify a ad r is to write out the first few terms: EXAMPLE 4 Is the series ò coverget or diverget? SOLUTION Let s rewrite the th term of the series i the form ar 2 : ò ò s2 2 d 2s2d ò 4 ò 4( 4 ) 2 2 We recogize this series as a geometric series with a 4 ad r 4. Sice r., the series diverges by (4). EXAMPLE 5 Write the umber as a ratio of itegers. SOLUTION After the first term we have a geometric series with a 7y0 ad r y0 2. Therefore EXAMPLE 6 Fid the sum of the series ò x, where x,. 0 SOLUTION Notice that this series starts with 0 ad so the first term is x 0. (With series, we adopt the covetio that x 0 eve whe x 0.) Thus ò x x x 2 x x 4 0 This is a geometric series with a ad r x. Sice r x ad (4) gives 5 EXAMPLE 7 Show that the series ò ò x 0 2 x s d,, it coverges is coverget, ad fid its sum. SOLUTION This is ot a geometric series, so we go back to the defiitio of a coverget series ad compute the partial sums. 206 Cegage Learig. All Rights Reserved. This cotet is ot yet fial ad Cegage Learig s o i isi d? 2 2?? 4 s d We ca simplify this expressio if we use the partial fractio decompositio isi d i 2 i

5 SERIES Cegage Learig. All Rights Reserved. This cotet is ot yet fial ad Cegage Learig Notice that the terms cacel i pairs. This is a example of a telescopig sum: Because of all the cacellatios, the sum collapses (like a pirate s collapsig telescope) ito just two terms. Figure illustrates Example 7 by showig the graphs of the sequece of terms a y[s d] ad the sequece hs j of partial sums. Notice that a l 0 ad s l. See Exer cises 76 ad 77 for two geometric iterpretatios of Example 7. 0 FIGURE s a The method used i Example 9 for showig that the harmoic series diverges is due to the Frech scholar Nicole Oresme (2 82). (see Sectio 5.6). Thus we have ad so s o i isi d i S o i 2 i D S 2 2D S 2 2 D S 2 4D S 2 D 2 lim s lim l` l`s D Therefore the give series is coverget ad ò EXAMPLE 8 Show that the harmoic series is diverget. s d ò 2 4 SOLUTION For this particular series it s coveiet to cosider the partial sums s 2, s 4, s 8, s 6, s 2,... ad show that they become large. s 2 2 s 4 2 s 4 d. 2 s 4 4 d 2 2 s 8 2 s 4 d s d. 2 s 4 4 d s d s 6 2 s 4 d s 5 8 d s 9 6 d. 2 s 4 4 d s 8 8 d s 6 6 d Similarly, s , s , ad i geeral s 2. 2 This shows that s 2 l ` as l ` ad so hs j is diverget. Therefore the harmoic series diverges. 6 Theorem If the series ò a is coverget, the lim l ` a 0.

6 6 SERIES PROOF Let s a a 2 a. The a s 2 s 2. Sice o a is coverget, the sequece hs j is coverget. Let lim l ` s s. Sice 2 l ` as l `, we also have lim l ` s 2 s. Therefore lim l ` a lim l `ss 2 s 2 d lim s 2 lim s 2 s 2 s 0 l ` l ` NOTE With ay series o a we associate two sequeces: the sequece hs j of its partial sums ad the sequece ha j of its terms. If o a is coverget, the the limit of the sequece hs j is s (the sum of the series) ad, as Theorem 6 asserts, the limit of the sequece ha j is 0. NOTE 2 The coverse of Theorem 6 is ot true i geeral. If lim l ` a 0, we caot coclude that o a is coverget. Observe that for the harmoic series oy we have a y l 0 as l `, but we showed i Example 8 that oy is diverget. 7 Test for Divergece If lim a does ot exist or if lim a ± 0, the the l` l` series ò a is diverget. The Test for Divergece follows from Theorem 6 because, if the series is ot diverget, the it is coverget, ad so lim l ` a 0. EXAMPLE 9 Show that the series ò SOLUTION diverges. 2 lim l` a lim l` lim l` So the series diverges by the Test for Divergece. 5 4y 2 5 ± 0 NOTE If we fid that lim l ` a ± 0, we kow that o a is diverget. If we fid that lim l ` a 0, we kow othig about the covergece or divergece of o a. Remember the warig i Note 2: if lim l ` a 0, the series o a might coverge or it might diverge. 8 Theorem If o a ad o b are coverget series, the so are the series o ca (where c is a costat), osa b d, ad osa 2 b d, ad (i) ò ca c ò a (ii) ò sa b d ò a ò b 206 Cegage Learig. All Rights Reserved. This cotet is ot yet fial ad Cegage Learig (iii) ò sa 2 b d ò a 2 ò b These properties of coverget series follow from the correspodig Limit Laws for Sequeces i Sectio 2.. For istace, here is how part (ii) of Theorem 8 is proved: Let s o t o a i i s ò a b i i t ò b

7 SERIES 7 The th partial sum for the series osa b d is ad thus we have u o i sa i b i d lim l ` u lim l ` o sa i b i d lim i l `So a i o b i i id lim l ` o a i lim i l ` o b i i lim l ` s lim l ` t s t 206 Cegage Learig. All Rights Reserved. This cotet is ot yet fial ad Cegage Learig Therefore o sa b d is coverget ad its sum is ò sa b d s t ò a ò b EXAMPLE 0 Fid the sum of the series S ò s d 2 D. SOLUTION The series oy2 is a geometric series with a 2 ad r 2, so I Example 7 we foud that ò ò s d So, by Theorem 8, the give series is coverget ad S D ò s d ò 2 s d ò 2? 4 NOTE 4 A fiite umber of terms does t affect the covergece or divergece of a series. For istace, suppose that we were able to show that the series is coverget. Sice ò ò ò 4 it follows that the etire series o ǹ ys d is coverget. Similarly, if it is kow that the series o ǹ N a coverges, the the full series is also coverget. ò a o N a ò a N

8 8 SERIES Exercises ;. (a) What is the differece betwee a sequece ad a series? (b) What is a coverget series? What is a diverget series? 2. Explai what it meas to say that o ǹ a 5. 4 Calculate the sum of the series o ǹ a whose partial sums are give.. s 2 2 s0.8d 4. s Calculate the first eight terms of the sequece of partial sums correct to four decimal places. Does it appear that the series is coverget or diverget? 5. ò 7. ò s 6. ò ls d s2d 8. ò 2! 9 4 Fid at least 0 partial sums of the series. Graph both the sequece of terms ad the sequece of partial sums o the same scree. Does it appear that the series is coverget or diverget? If it is coverget, fid the sum. If it is diverget, explai why ò s25d 0. ò cos. ò s 2 4. S ò s 2 s D 7 2. ò 0 4. ò 2 s 2d 5. Let a 2. (a) Determie whether ha j is coverget. (b) Determie whether o ǹ a is coverget. 6. (a) Explai the differece betwee o a i ad o a j i j (b) Explai the differece betwee o a i ad o a j i i 7 26 Determie whether the geometric series is coverget or diverget. If it is coverget, fid its sum ò 6s0.9d ò s29d 2 s2d 2. ò ò 0 s s2 d 25. ò ò e Determie whether the series is coverget or diverget. If it is coverget, fid its sum ò 2 ksk 2d 0. ò k sk d 2 2. ò 2. ò 2. ò s 2 5. ò 7. ò ls 2 k 0S D k 2 2 D 9. ò arcta 4. S ò e s dd 4. ò fs0.8d 2 2 s0.d g 6. ò s 2 d 8. ò scos d k k 40. S ò 5 e 42. ò 2 2 D 4 48 Determie whether the series is coverget or diverget by expressig s as a telescopig sum (as i Ex am ple 7). If it is coverget, fid its sum ò ò s d 46. ò Scos 2 2 cos 47. ò se y 2 e ysd d 2D s d 44. ò l 48. ò Cegage Learig. All Rights Reserved. This cotet is ot yet fial ad Cegage Learig

9 SERIES Cegage Learig. All Rights Reserved. This cotet is ot yet fial ad Cegage Learig CAS 49. Let x (a) Do you thik that x, or x? (b) Sum a geometric series to fid the value of x. (c) How may decimal represetatios does the umber have? (d) Which umbers have more tha oe decimal represetatio? 50. A sequece of terms is defied by a Calculate o ǹ a. a s5 2 da Express the umber as a ratio of itegers Fid the values of x for which the series coverges. Fid the sum of the series for those values of x. 57. ò s25d x 58. ò sx 2d sx 2 2d 59. ò 60. ò s24d sx 2 5d ò 0 x 6. ò e x si ò x We have see that the harmoic series is a diverget series whose terms approach 0. Show that ls D ò is aother series with this property Use the partial fractio commad o your CAS to fid a coveiet expressio for the partial sum, ad the use this expressio to fid the sum of the series. Check your aswer by usig the CAS to sum the series directly. 65. ò ò s 2 d 67. If the th partial sum of a series o ǹ a is fid a ad o ǹ a. s If the th partial sum of a series oǹ a is s 2 22, fid a ad o ǹ a. 69. Drug dosig A patiet takes 50 mg of a drug at the same time every day. Just before each tablet is take, 5% of the drug remais i the body. (a) What quatity of the drug is i the body after the third tablet? After the th tablet? (b) What quatity of the drug remais i the body i the log ru? 70. Isuli ijectio After ijectio of a dose D of isuli, the cocetratio of isuli i a patiet s system decays expoetially ad so it ca be writte as De 2at, where t represets time i hours ad a is a positive costat. (a) If a dose D is ijected every T hours, write a expressio for the sum of the residual cocetratios just before the s dst ijectio. (b) Determie the limitig pre-ijectio cocetratio. (c) If the cocetratio of isuli must always remai at or above a critical value C, determie a miimal dosage D i terms of C, a, ad T. 7. Whe moey is spet o goods ad services, those who receive the moey also sped some of it. The people receivig some of the twice-spet moey will sped some of that, ad so o. Ecoomists call this chai reactio the multiplier effect. I a hypothetical isolated commuity, the local govermet begis the process by spedig D dollars. Suppose that each recipiet of spet moey speds 00c% ad saves 00s% of the moey that he or she receives. The val ues c ad s are called the margial propesity to cosume ad the margial propesity to save ad, of course, c s. (a) Let S be the total spedig that has bee geerated after trasactios. Fid a equatio for S. (b) Show that lim l ` S kd, where k ys. The umber k is called the multiplier. What is the multiplier if the margial propesity to cosume is 80%? Note: The federal govermet uses this priciple to justify deficit spedig. Baks use this priciple to justify led ig a large percetage of the moey that they receive i deposits. 72. A certai ball has the property that each time it falls from a height h oto a hard, level surface, it rebouds to a height rh, where 0, r,. Suppose that the ball is dropped from a iitial height of H meters. (a) Assumig that the ball cotiues to bouce idefiitely, fid the total distace that it travels. (b) Calculate the total time that the ball travels. (Use the fact that the ball falls 2 tt 2 meters i t secods.) (c) Suppose that each time the ball strikes the surface with velocity v it rebouds with velocity 2kv, where 0, k,. How log will it take for the ball to come to rest? 7. Fid the value of c if 74. Fid the value of c such that ò s cd ò e c 0 0

10 0 SERIES ; 75. I Example 8 we showed that the harmoic series is diverget. Here we outlie aother method, makig use of the fact that e x. x for ay x. 0. If s is the th partial sum of the harmoic series, show that e s.. Why does this imply that the harmoic series is diverget? 76. Graph the curves y x, 0 < x <, for 0,, 2,, 4,... o a commo scree. By fidig the areas betwee successive curves, give a geometric demostratio of the fact, show i Example 7, that ò s d 77. The figure shows two circles C ad D of radius that touch at P. The lie T is a commo taget lie; C is the circle that touches C, D, ad T; C 2 is the circle that touches C, D, ad C ; C is the circle that touches C, D, ad C 2. This procedure ca be cotiued idefiitely ad produces a ifiite sequece of circles hc j. Fid a expressio for the diameter of C ad thus provide aother geometric demostratio of Example 7. C C C 78. A right triagle ABC is give with /A ad AC b. CD is draw perpedicular to AB, DE is draw perpedicular to BC, EF AB, ad this process is cotiued idefiitely, as show i the figure. Fid the total legth of all the perpediculars P C CD DE EF FG i terms of b ad. B H F G D E C A b D T 79. What is wrog with the followig calculatio? s 2 d s 2 d s 2 d s2 d s2 d s2 d (Guido Ubaldus thought that this proved the existece of God because somethig has bee created out of othig. ) 80. Suppose that o ǹ a sa ± 0d is kow to be a coverget series. Prove that oǹ ya is a diverget series. 8. Prove part (i) of Theorem If o a is diverget ad c ± 0, show that o ca is diverget. 8. If o a is coverget ad o b is diverget, show that the series o sa b d is diverget. [Hit: Argue by cotradictio.] 84. If o a ad o b are both diverget, is o sa b d ecessarily diverget? 85. Suppose that a series o a has positive terms ad its partial sums s satisfy the iequality s < 000 for all. Explai why o a must be coverget. 86. The Fiboacci sequece is defied by the equatios f, f 2, f f 2 f 22 > Show that each of the followig statemets is true. (a) (b) ò 2 2 f 2 f f 2 f f f f 2 f (c) ò 2 f f 2 f The Cator set, amed after the Germa mathematicia Georg Cator (845 98), is costructed as follows. We start with the closed iterval [0, ] ad remove the ope iterval (, 2 ). That leaves the two itervals f0, g ad f 2, g ad we remove the ope middle third of each. Four itervals remai ad agai we remove the ope middle third of each of them. We cotiue this procedure idefiitely, at each step removig the ope middle third of every iterval that remais from the precedig step. The Cator set cosists of the umbers that remai i [0, ] after all those itervals have bee removed. (a) Show that the total legth of all the itervals that are removed is. Despite that, the Cator set cotais ifiitely may umbers. Give examples of some umbers i the Cator set. (b) The Sierpiski carpet is a two-dimesioal couterpart of the Cator set. It is costructed by removig the ceter oe-ith of a square of side, the removig the 206 Cegage Learig. All Rights Reserved. This cotet is ot yet fial ad Cegage Learig

11 SERIES ceters of the eight smaller remaiig squares, ad so o. (The figure shows the first three steps of the costructio.) Show that the sum of the areas of the removed squares is. This implies that the Sierpiski carpet has area 0. (b) Use mathematical iductio to prove your guess. (c) Show that the give ifiite series is coverget, ad fid its sum. 90. I the figure at the right there are ifiitely may circles approachig the vertices of a equilateral triagle, each circle touchig other circles ad sides of the triagle. If the triagle has sides of legth, fid the total area occupied by the circles. 206 Cegage Learig. All Rights Reserved. This cotet is ot yet fial ad Cegage Learig 88. (a) A sequece ha j is defied recursively by the equatio a 2 sa2 a22d for >, where a ad a2 ca be ay real umbers. Experimet with various values of a ad a 2 ad use your calculator to guess the limit of the sequece. (b) Fid lim l ` a i terms of a ad a 2 by expressig a 2 a i terms of a 2 2 a ad summig a series. 89. Cosider the series o ǹ ys d!. (a) Fid the partial sums s, s 2, s, ad s 4. Do you recogize the deomiators? Use the patter to guess a formula for s.

12 2 SERIES Aswers. (a) A sequece is a ordered list of umbers whereas a series is the sum of a list of umbers. (b) A series is coverget if the sequece of partial sums is a coverget sequece. A series is diverget if it is ot coverget ,.25,.620,.777,.857,.90,.92,.952; C ,.284, ,.7598, 5.049, 7.044, ,.0540; D , , , , sa d , , , , , ; ss d coverget, sum ,.542,.9867, ,.80927, , , , , ; diverget , , , , , , , , , ; coverget, sum _ 0 0 ss d sa d 0 ss d sa d 5. (a) Yes (b) No 7. D D 27. D 29. D D 5. D 7. D 9. D 4. eyse 2 d e (b) (c) 2 (d) All ratioal umbers with a termiatig decimal represetatio, except y , x, 5 ; 25x 5x 59. 2, x, 5; 5 2 x 6. x. 2 or x, 22; x x a 0, a 6. x, 0; 2 e x 2 for., sum s d (a) mg; 9 s d (b) mg 7. (a) S Ds 2 c d (b) ss 2 d 2 c The series is diverget. s d 85. hs j is bouded ad icreasig. 87. (a) 0, 9, 2 9,, 2, 7 9, 8 9, 89. (a) 2, 5 6, 2 24, 9 20 ; s d! 2 s d! (c) 206 Cegage Learig. All Rights Reserved. This cotet is ot yet fial ad Cegage Learig

13 SERIES Solutios. (a) A sequece is a ordered list of umbers whereas a series is the sum of a list of umbers.. (b) A series is coverget if the sequece of partial sums is a coverget sequece. A series is diverget if it is ot coverget. = 5. For = lim = lim [2 (08) ]= lim 2 lim (08) =2 (0) = 2 =, =. = = =, 2 = + 2 =+ =25, = , Cegage Learig. All Rights Reserved. This cotet is ot yet fial ad Cegage Learig 4 = , 5 = , 6 = , 7 = ,ad 8 = It appears that the series is coverget. 7. For = +, = +. = = + =05, 2 2 = + 2 = , = , 4 = , 5 = , 6 = , 9. 7 = , 8 = It appears that the series is diverget From the graph ad the table, it seems that the series coverges to 2. I fact, it is a geometric series with = 24 ad =,soitssumis 2 5 = ( 5) = 24 = 24 2 = 2 5 Note that the dot correspodig to =is part of both { } ad { }. TI-86 Note: To graph { } ad { }, set your calculator to Param mode ad DrawDot mode. (DrawDot is uder GRAPH, MORE, FORMT (F).) Now uder E(t)= make the assigmets: xt=t, yt=2/(-5)ˆt, xt2=t, yt2=sum seq(yt,t,,t,). (sum ad seq are uder LIST, OPS (F5), MORE.) Uder WIND use,0,,0,0,,-,, to obtai a graph similar to the oe above. The use TRACE (F4) to see the values.

14 4 SERIES The series = diverges, sice its terms do ot approach From the graph ad the table, it seems that the series coverges. = + = so = = 2, + = lim =. 5. (a) lim 2 = lim + = 2,sothesequece { } is coverget by (2..). (b) Sice lim = 2 6=0,theseries is diverget by the Test for Divergece. = is a geometric series with ratio = 4.Sice = 4, the series diverges is a geometric series with ratio 2 =.Sice = 0 5 5, the series coverges to = 0 ( 5) = 0 65 = 50 6 = Cegage Learig. All Rights Reserved. This cotet is ot yet fial ad Cegage Learig 2. 6(09) is a geometric series with first term =6ad ratio =09. Sice =09, the series coverges to = = 6 09 = 6 0 =60.

15 SERIES = ( ) coverges to =0 4 = 4 + =. The latter series is geometric with =ad ratio = 4 4.Sice =, it 4 = ( 4) = 4. Thus, the give series coverges to =. 7 = = it diverges. is a geometric series with ratio =.Sice, the series diverges. = = =. This is a costat multiple of the diverget harmoic series, so 206 Cegage Learig. All Rights Reserved. This cotet is ot yet fial ad Cegage Learig 29. = diverges by the Test for Divergece sice lim = lim = 6=0.. Coverges = = +2 = = + 2 = = + = = 2 +2= = diverges by the Test for Divergece sice lim = lim 2 = lim 2 =2 0 =6= 0. = 2 + l diverges by the Test for Divergece sice lim = lim l =l lim =l =0. =0 = = is a geometric series with ratio = 047. It diverges because. arcta diverges by the Test for Divergece sice lim = lim arcta = 2 6=0. [sum of two coverget geometric series] = is a geometric series with first term = = ad ratio =.Sice =, the series coverges to = = =.ByExample7, + = ( +) = + = = =. Thus, by Theorem 8(ii), ( +) ( +) = += + =.

16 6 SERIES 4. Usig partial fractios, the partial sums of the series =2 = = =2 =2 2 2 are 2 ( )( +) = = This sum is a telescopig series ad = Thus, 2 = lim = lim + 2 = For the series = ( +), = = ( +) = = Thus, = 47. For the series [usig partial fractios]. The latter sum is = [telescopig series] ( +) = lim = lim + + =+ + =. Coverges = (+), = (+) = =( 2 )+( 2 )+ + (+) = (+) + [telescopig series] Thus, (+) = lim = lim (+) = 0 =. Coverges = 49. (a) May people would guess that, but ote that cosists of a ifiite umber of 9s. (b) = = ,000 + = =0. Itssumis 09 0 = 09 =,thatis, =. 09 (c) The umber has two decimal represetatios, ad = 9 0, which is a geometric series with =09 ad (d) Except for 0, all ratioal umbers that have a termiatig decimal represetatio ca be writte i more tha oe way. For example, 05 ca be writte as as well as Cegage Learig. All Rights Reserved. This cotet is ot yet fial ad Cegage Learig = is a geometric series with = 02 0 ad = 0.Itcovergesto = 80 0 = = Now is a geometric series with = 06 0 ad = 0.Itcovergesto = = = Thus, 256 = = = 88.

17 SERIES = Now is a geometric series with = ad = It coverges to = = = Thus, 542 = = = 5, = 5, or ( 5) = ( 5) is a geometric series with = 5, so the series coverges = = 5,thatis,. I that case, the sum of the series is = 5 ( 5) = Cegage Learig. All Rights Reserved. This cotet is ot yet fial ad Cegage Learig ( 2) 59. = 2 is a geometric series with = 2, so the series coverges =0 = I that case, the sum of the series is = 2 =0 2 = =0 2 = ( 2) = 5. is a geometric series with = 2, so the series coverges 2 2 or 2. I that case, the sum of the series is = 2 = 2. 2 = ( ) is a geometric series with =, so the series coverges =0 = I that case, the sum of the series is =. 65. After defiig, Weusecovert(f,parfrac); i Maple, Apart i Mathematica, or Expad Ratioal ad Simplify i Derive to fid that the geeral term is = ( 2 + ).Sotheth partial sum is ( +) = = ( +) = ( +) = ( +) The series coverges to lim =. This ca be cofirmed by directly computig the sum usig sum(f,=..ifiity); (i Maple), Sum[f,{,,Ifiity}] (i Mathematica), or Calculus Sum (from to )adsimplify (i Derive). 67. For =, =0sice =0.For, = = ( ) ( ) ( +)( 2) 2 = = + ( ) + ( +) ( +) Also, = = lim = lim + =.

18 8 SERIES 69. (a) The quatity of the drug i the body after the first tablet is 50 mg. After the secod tablet, there is 50 mg plus 5% of the first 50- mg tablet, that is, [ (005)] mg. After the third tablet, the quatity is [ (005) + 50(005) 2 ]=57875 mg. After tablets, the quatity (i mg) is (005) + +50(005). We ca use Formula to write this as 50( 005 ) 005 = ( 005 ). (b) The umber of milligrams remaiig i the body i the log ru is lim 000 ( ) = 000 ( 0) 57895, 9 oly 002 mg more tha the amout after tablets. 7. (a) The first step i the chai occurs whe the local govermet speds dollars. The people who receive it sped a fractio of those dollars, that is, dollars. Those who receive the dollars sped a fractio of it, that is, 7. 2 dollars. Cotiuig i this way, we see that the total spedig after trasactios is = = ( ) by (). (b) lim ( ) = lim = lim ( )= = [sice + =] = [sice =] If =08,the = =02adthe multiplier is = =5. sice 0 lim =0 ( + ) is a geometric series with =(+) 2 ad =(+), so the series coverges whe =2 ( + ) + + or + 0 or 2. We calculate the sum of the series ad set it equal to 2: ( + ) 2 =2 ( + ) 2 = =2(+) 2 2( + ) =0 = 2 ± 2 4 = ± 2. However, the egative root is iadmissible because So = = = 2 ( + ) [ +] = = + 2 Thus, +ad lim =. Sice{ } is icreasig, lim =, implyig that the harmoic series is 206 Cegage Learig. All Rights Reserved. This cotet is ot yet fial ad Cegage Learig diverget.

19 SERIES Let be the diameter of. We draw lies from the ceters of the to the ceter of (or ), ad usig the Pythagorea Theorem, we ca write = = =2 [differece of squares] =. 2 Similarly, = = =(2 )( + 2 ) 2 = ( )2 =, = = [ ( + 2)]2, ad i geeral, 2 ( + 2) 206 Cegage Learig. All Rights Reserved. This cotet is ot yet fial ad Cegage Learig + = = 2 2 =. If we actually calculate 2 ad from the formulas above, we fid that they are 6 = 2 ad 2 = 4 respectively, so we suspect that i geeral, =. To prove this, we use iductio: Assume that for all ( +), = ( +) = +.The = 2 + formula for +,weget + = = 2 + = + = ( +) = + [telescopig sum]. Substitutig this ito our, ad the iductio is complete. ( +)( +2) Now, we observe that the partial sums = of the diameters of the circles approach as ;thatis, = = = =, which is what we wated to prove. ( +) 79. The series diverges (geometric series with = ) so we caot say that 0= = = lim = = lim = = lim = = =, which exists by hypothesis. 8. Suppose o the cotrary that ( + ) coverges. The ( + ) ad are coverget series. So by Theorem 8(iii), [( + ) ] would also be coverget. But [( + ) ]=, a cotradictio, sice is give to be diverget. 85. The partial sums { } form a icreasig sequece, sice = 0 for all. Also, the sequece { } is bouded sice 000 for all. So by the Mootoic Sequece Theorem, the sequece of partial sums coverges, that is, the series is coverget.

20 20 SERIES 87. (a) At the first step, oly the iterval 2 (legth ) is removed. At the secod step, we remove the itervals ad , which have a total legth of 2 2. At the third step, we remove 2 2 itervals, each of legth. I geeral, at the th step we remove 2 itervals, each of legth,foralegthof2 = 2. Thus, the total legth of all removed itervals is 2 = = 2 geometric series with = ad = 2. Notice that at = the th step, the leftmost iterval that is removed is 2,soweeverremove0,ad0 is i the Cator set. Also, the rightmost iterval removed is 2,sois ever removed. Some other umbers i the Cator set are, 2, 9, 2 9, 7 9,ad 8 9. (b) The area removed at the first step is ;atthesecodstep, ;atthethirdstep,(8) 2 9. I geeral, the area removed at the th step is (8) = 8, so the total area of all removed squares is = =. = 89. (a) For = ( +)!, = 2 = 2, 2 = = 5 6, = = 2 24, 4 = = 9 20 ( +)!. The deomiators are ( +)!,soaguesswouldbe =. ( +)! (b) For =, = 2 = 2!, so the formula holds for =. Assume = 2! + = = ( +)! ( +)! ( +2)! ( +2)! ( +)!.The ( +)! + + ( +)! + ( +2)! ( +2)+ + = + = ( +2)! ( +)! ( +)!( +2) ( +2)! Thus, the formula is true for = +. So by iductio, the guess is correct. (c) lim ( +)! = lim = lim ( +)! =ad so ( +)! = ( +)! =. 206 Cegage Learig. All Rights Reserved. This cotet is ot yet fial ad Cegage Learig

21 206 Cegage Learig. All Rights Reserved. This cotet is ot yet fial ad Cegage Learig SERIES 2

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28 28 SERIES 206 Cegage Learig. All Rights Reserved. This cotet is ot yet fial ad Cegage Learig