10.2 Infinite Series Contemporary Calculus 1

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1 10. Ifiite Series Cotemporary Calculus INFINITE SERIES Our goal i this sectio is to add together the umbers i a sequece. Sice it would take a very log time to add together the ifiite umber of umbers, we first cosider fiite sums, look for patters i these fiite sums, ad take its as more ad more umbers are icluded i the fiite sums. What does it mea to add together a ifiite umber of terms? We will defie that cocept carefully i this sectio. Secodly, is the sum of all the terms a fiite umber? I the ext few sectios we will examie a variety of techiques for determiig whether a ifiite sum is fiite. Fially, if we kow the sum is fiite, ca we determie the value of the sum? The difficulty of fidig the exact value of the sum varies from very easy to very, very difficult. Example 1: A golf ball is throw 9 feet straight up ito the air, ad o each bouce it rebouds to two thirds of its previous height (Fig. 1). Fid a sequece whose terms give the distaces the ball travels durig each successive bouce. Represet the total distace traveled by the ball as a sum. Solutio: The heights of the successive bouces are 9 feet, ( 3 ). 9 feet, ( 3 ). [( 3 ). 9] feet, ( 3 )3. 9 feet, ad so forth. O each bouce, the ball rises ad falls so the distace traveled is twice the height of that bouce: 18 feet, ( 3 ). 18 feet, ( 3 ). ( 3 ). 18 feet, ( 3 )3. 18 feet, ( 3 )4. 18 feet,.... The total distace traveled is the sum of the bouce distaces: total distace = 18 + ( 3 ) ( 3 ). ( 3 ) ( 3 ) ( 3 ) = 18 { ( 3 ) + ( 3 )3 + ( 3 ) } At the completio of the first bouce the ball has traveled 18 feet. After the secod bouce, it has traveled 30 feet, a total of 38 feet after the third bouce, feet after the fourth, ad so o. With a calculator ad some patiece, we see that after the 0 th bouce the ball has traveled a total of approximately feet, after the 30 th bouce approximately feet, ad after the 40th bouce approximately feet.

2 10. Ifiite Series Cotemporary Calculus Practice 1: A teis ball is throw 10 feet straight up ito the air, ad o each bouce it rebouds to 40% of its previous height. Represet the total distace traveled by the ball as a sum, ad fid the total distace traveled by the ball after the completio of its third bouce. (Fig. ) Ifiite Series The ifiite sums i the Example ad Practice are called ifiite series, ad they are the objects we will start to examie i this sectio. Defiitios A ifiite series is a expressio of the form a 1 + a + a 3 + a or ak. The umbers a 1, a, a 3, a 4,... are called the terms of the series. (Fig. 3) Example : Represet the followig series usig the sigma otatio. (a) 1 + 1/3 + 1/9 + 1/7 +..., (b) 1 + 1/ 1/3 + 1/4 1/5 +..., (c) 18( /3 + 4/9 + 8/7 + 16/ ) (d) = 7/ / / , ad (e) 0... Solutio: (a) 1 + 1/3 + 1/9 + 1/ = ( 1 3 ) k or ( 1 3 ) k 1 k=0 (b) 1 + 1/ 1/3 + 1/4 1/ = ( 1) k 1 k (c) 18 ( 3 ) k (d) = 7/ / / = 7 10 k (e) 10 k Practice : Represet the followig series usig the sigma otatio. (a) , (b) (c) / + 1/ (d) 1/ + 1/4 + 1/6 + 1/8 + 1/ (e)

3 10. Ifiite Series Cotemporary Calculus 3 I order to determie if the ifiite series adds up to a fiite value, we examie the sums as more ad more terms are added. Defiitio The partial sums s of the ifiite series are the umbers s 1 = a 1, s = a 1 + a, s 3 = a 1 + a + a 3,... ak I geeral, s = a 1 + a + a a = ak or, recursively, as s = s 1 + a. The partial sums form the sequece of partial sums { s }. Example 3: Calculate the first 4 partial sums for the followig series. (a) 1 + 1/ + 1/4 + 1/8 + 1/ , (b) ( 1) k, ad (c) 1. Solutio: (a) s 1 = 1, s = 1 + 1/ = 3/, s 3 = 1 + 1/ + 1/4 = 7/4, s 4 = 1 + 1/ + 1/4 + 1/8 = 15/8 It is usually easier to use the recursive versio of s : s 3 = s + a 3 = 3/ + 1/4 = 7/4 ad s 4 = s 3 + a 4 = 7/4 + 1/8 = 15/8. (b) s 1 = ( 1) 1 = 1, s = s 1 + a = 1 + ( 1) = 0, s 3 = s + a 3 = 0 + ( 1) 3 = 1, s 4 = 0. (c) s 1 = 1, s = 3/, s 3 = 11/6, s 4 = 5/1. Practice 3: Calculate the first 4 partial sums for the followig series. (a) 1 1/ + 1/4 1/8 + 1/16..., (b) ( 1 3 ) k, ad (c) ( 1). = If we kow the values of the partial sums s, we ca recover the values of the terms a used to build the s. Example 4: Suppose s 1 =.1, s =.6, s 3 =.84, ad s 4 =.87 are the first partial sums of ak. Fid the values of the first four terms of { a }. Solutio: s 1 = a 1 so a 1 =.1. s = a 1 + a so.6 =.1 + a ad a = 0.5. Similarly, s 3 = a 1 + a + a 3 so.84 = a 3 ad a 3 = 0.4. Fially, a 4 = 0.03.

4 10. Ifiite Series Cotemporary Calculus 4 A alterate solutio method starts with a 1 = s 1 ad the uses the fact that s = s 1 + a so a = s s 1. The a = s s 1 =.6.1 = 0.5. a 3 = s 3 s =.84.6 = 0.4, ad a 4 = s 4 s 3 = = Practice 4: Example 5: Suppose s 1 = 3., s = 3.6, s 3 = 3.5, s 4 = 4, s 99 = 7.3, s 100 = 7.6, ad s 101 = 7.8 are partial sums of ak. Fid the values of a 1, a, a 3, a 4, ad a 100. Graph the first five terms of the series ( 1 ) k. The graph the first five partial sums. Solutio: a 1 = $ 1% 1 ' # & = 1, a = $ 1% # ' & = 1 4, a3 = $ 1% # & 3 ' = 1, a4 = , a 4 = 1 3 s 1 = 1, s = = 3 4, s 3 = = 7 8, s 4 = 15 16, ad s 5 = These values are graphed i Fig. 4. Practice 5: Graph the first five terms of the series ( 1 ) k. The graph the first five partial sums. Covergece of a Series The covergece of a series is defied i terms of the behavior of the sequece of partial sums. If the partial sums, the sequece obtaied by addig more ad more of the terms of the series, approaches a fiite umber, we say the series coverges to that fiite umber. If the sequece of partial sums diverges (does ot approach a sigle fiite umber), we say that the series diverges.

5 10. Ifiite Series Cotemporary Calculus 5 Defiitios Let { s } be the sequece of partial sums of the series ak : s = ak If { s } is a coverget sequece, (Fig. 5) we say the series ak coverges. If the sequece of partial sums { s } coverges to A, we say the series ak coverges to A or the sum of the series is A, ad we write ak = A. If the sequece of partial sums { s } diverges, we say the series ak diverges. Example 6: Solutio: I the ext sectio we preset a method for determiig that the th partial sum of ( 1 ) k is s = 1 = 1 1. Use this result to evaluate the it of { s }. Does the series # s = Practice 6: The th partial sum of ( 1 ) k coverge? If so, to what value? # 1 1 = 1 (Fig. 6), so evaluate the it of { s }. Does the series ( 1 ) k coverges to 1: ( 1 ) k = 1. ( 1 ) k is s = ( 1 ). Use this result to ( 1 ) k coverge? If so, to what value?

6 10. Ifiite Series Cotemporary Calculus 6 The ext theorem says that if a series coverges, the the terms of the series must approach 0. Whe a series is coverget the the partial sums s approach a fiite it (Fig. 3) so all of the s must be close to that it whe is large. The s ad s 1 must be close to each other (why?) ad a = s s 1 must be close to 0. Theorem: If the series ak coverges, the k# a k = 0. We ca NOT use this theorem to coclude that a series coverges. If the terms of the series do approach 0, the the series may or may ot coverge more iformatio is eeded to draw a coclusio. However, a alterate form of the theorem, called the th Term Test for Divergece, is very useful for quickly cocludig that some series diverge. th Term Test for Divergece of a Series If the the terms a of a series do ot approach 0 (as ) the series diverges: if # a 0, the a diverges. Example 7: Solutio: Which of these series diverge by the th Term Test? (a) ( 1 ) (b) ( 3 4 ) (c) ( ) (d) 1 (a) a = ( 1 ) oscillates betwee 1 ad +1 ad does ot approach 0. ( 1 ) diverges. (b) a = ( 3 4 ) approaches 0 so ( 3 4 ) may or may ot coverge. (c) a = ( ) approaches e 0, so ( ) diverges. (d) a = 1 approaches 0 so 1 may or may ot coverge. We ca be certai that (a) ad (c) diverge. We do't have eough iformatio yet to decide about (b) ad (d). (I the ext sectio we show that (b) coverges ad (d) diverges.) Practice 7: Which of these series diverge by the th Term Test?

7 10. Ifiite Series Cotemporary Calculus 7 (a) ( 0.9 ) (b) ( 1.1 ) (c) si( π ) (d) 1 New Series From Old If we kow about the covergece of a series, the we also kow about the covergece of several related series. Isertig or deletig a few terms, ay fiite umber of terms, does ot chage the covergece or divergece of a series. The isertios or deletios typically chage the sum (the it of the partial sums), but they do ot chage whether or ot the series coverges. (Isertig or deletig a ifiite umber of terms ca chage the covergece or divergece.) Multiplyig each term i a series by a ozero costat does ot chage the covergece or divergece of a series: Suppose c 0. a coverges if ad oly if c. a coverges. Term by term additio ad subtractio of the terms of two coverget series result i coverget series. (Term by term multiplicatio ad divisio of series do ot have such ice results.) Theorem: If a ad b coverge with a = A ad b = B, the C. a = C. A, (a + b ) = A + B, ad (a b ) = A B. The proofs of these statemets follow directly from the defiitio of covergece of a series ad from results about covergece of sequeces (of partial sums).

8 10. Ifiite Series Cotemporary Calculus 8 PROBLEMS I problems 1 6, rewrite each sum usig sigma otatio startig with k = si(1) + si(8) + si(7) + si(64) + si(15) ( 1 ) + ( 1 4 ) + ( 1 8 ) + ( 1 16 ) + ( 1 3 ) ( 1 3 ) + ( 1 9 ) + ( 1 7 ) + ( 1 81 ) + ( 1 43 ) +... I problems 7 14, calculate ad graph the first four partial sums s 1 to s 4 of the give series ( 1) a. 10. { } ( ) I problems 13 18, the first five partial sums s 1 to s 5 are give. Fid the first four terms a 1 to a 4 of the series. 13. s 1 = 3, s =, s 3 = 4, s 4 = 5, s 5 = s 1 = 3, s = 5, s 3 = 4, s 4 = 6, s 5 = s 1 = 4, s = 4.5, s 3 = 4.3, s 4 = 4.8, s 5 = s 1 = 4, s = 3.7, s 3 = 3.9, s 4 = 4.1, s 5 = s 1 = 1, s = 1.1, s 3 = 1.11, s 4 = 1.111, s 5 = s 1 = 1, s = 0.9, s 3 = 0.93, s 4 = 0.91, s 5 = 0.9 I problems 19 8, represet each repeatig decimal as a series usig the sigma otatio aaa ababab Fid a patter for a fractio represetatio of the repeatig decimal 0.abcabcabc A golf ball is throw 0 feet straight up ito the air, ad o each bouce it rebouds to 60% of its previous height. Represet the total distace traveled by the ball as a sum. 31. A super ball is throw 15 feet straight up ito the air, ad o each bouce it rebouds to 80% of its previous height. Represet the total distace traveled by the ball as a sum.

9 10. Ifiite Series Cotemporary Calculus 9 3. Each special washig of a pair of overalls removes 80% of the radioactive particles attached to the overalls. Represet, as a sequece of umbers, the percet of the origial radioactive particles that remai after each washig. 33. Each week, 0% of the argo gas i a cotaier leaks out of the cotaier. Represet, as a sequece of umbers, the percet of the origial argo gas that remais i the cotaier at the ed of the 1 st, d, 3 rd, ad th weeks. 34. Eight people are goig o a expeditio by horseback through desolate coutry. The people ad scietific equipmet (fishig gear) require 1 horses, ad additioal horses are eeded to carry food for the horses. Each horse ca carry eough food to feed horses for the trip. Represet the umber of horses eeded to carry food as a sum. (Start of a solutio: The origial 1 horses will require 6 ew horses to carry their food. The 6 ew horses require 3 additioal horses to carry their food. The 3 additioal horses require aother 1.5 horses to carry food for them, etc. ) Which of the series i problems 35 43, defiitely diverge by the th Term Test? What ca we coclude about the other series i these problems? 35. ( ) 1 # # 37. # 4 ( 3) ( ) # 38. $ si() # 40. # l() ( ) 41. # cos 1 4. # 0 $ # $ Practice Aswers Practice 1: The heights of the bouces are 10, (0.4). 10, (0.4). (0.4). 10, (0.4) 3. 10,... so the distaces traveled (up ad dow) by the ball are 0, (0.4). 0, (0.4). (0.4). 0, (0.4) 3. 0,... The total distace traveled is 0 + (0.4). 0 + (0.4). 0 + (0.4) = 0{ (0.4) + (0.4) } = 0 (0.4) k k=0 After 3 bouces the ball has traveled 0 + (0.4)(0) + (0.4) (0) = = 31. feet. Practice : (a) k (b) ( 1) k

10 10. Ifiite Series Cotemporary Calculus 10 (c) ( ) = ( 1 k=0 ) k or ( 1 k=0 ) k 1 or ( 1 ) k (d) 1 k (e) = = ( 1 10 ) k or 1 10 k Practice 3: (a) Partial sums: 1, 1/, 3/4, 5/8 (b) ; partial sums: 1 3, 4 9, 13 7, (c) ; partial sums: 1, 1 6, 10 4 = 5 1, Practice 4: a 1 = s 1 = 3., a = s s 1 = (3.6) (3.) = 0.4, a 3 = s 3 s = (3.5) (3.6) = 0.1, a 4 = s 4 s 3 = (4) (3.5) = 0.5, a 100 = s 100 s 99 = (7.6) (7.3) = 0.3 Practice 5: a 1 = 1/, a = 1/4, a 3 = 1/8, a 4 = 1/16, a 5 = 1/3 s 1 = a 1 = 1, s = a 1 + a = = 1 4, 0.5 Practice 6: s 3 = s + a 3 = = , s 4 = s 3 + a 4 = = , s 5 = s 4 + a 5 = = , The graphs of a ad s are show i Fig. 6. # s = = 1 3. # ( 1 ) = The it, as, of s is a fiite umber, so a Fig. 6 ( 1 ) k coverges to 1 3. s Practice 7: (a) ( 0.9 ) : a = ( 0.9 ) approaches 0 so ( 0.9 ) may coverge. (b) ( 1.1 ) : a = ( 1.1 ) approaches ifiity so ( 1.1 ) diverges. (c) si( π ) = : si( π ) = 0 approaches 0 so si( π ) may coverge. (d) 1 : a = 1 approaches 0 so 1 may coverge. (Later i this chapter we will show that series (a) ad (c) coverge ad series (d) diverges.)

11 10. Ifiite Series Cotemporary Calculus 11 Appedix: Programmig Partial Sums of Numerical Series MAPLE commads for 1001, ,! : > sum(1/,..100) ; (the press ENTER key) > sum(1/^,..100) ; > sum(1/(!),..100) ; TI 85 program for M 1 Prgm1:NUMSUM Disp NUM TERMS = Iput M 1 A 0 N Lbl ONE N+1 N 1/N A (value of the ew term) 1/N A S + A S (add ew term to partial sum) S + A S If N<M (test if eed ext term) If N<M Goto ONE Disp S Stop Stop M 1 For M 1 chage the bold lie to 1/(N*N) A. For! chage the bold lie to A/N A.