10.2 Series Contemporary Calculus 110.2 SERIESOur goal in this section is to add together the numbers in a sequence. Since it would take a "very long time"to add together the infinite number of numbers, we first consider finite sums, look for patterns in these finitesums, and take limits as more and more numbers are included in the finite sums.What does it mean to add together an infinite number of terms? We will define that concept carefully inthis section. Secondly, is the sum of all the terms a finite number? In the next few sections we willexamine a variety of techniques for determining whether an infinite sum is finite. Finally, if we know thesum is finite, can we determine the value of the sum. The difficulty of finding the exact value of the sumvaries from very easy to very, very difficult.Example 1:A golf ball is thrown 9 feet straight up intothe air, and on each bounce it rebounds to two thirds ofits previous height (Fig. 1). Find a sequence whoseterms give the distances the ball travels during eachsuccessive bounce. Represent the total distancetraveled by the ball as a sum.Solution: The heights of the successive bounces are 9 feet,( 23 ).9 feet, ( 23 ).[( 23 ).9] feet, ( 23 )3 .9 feet, andso forth. On each bounce, the ball rises and falls so thedistance travelled is twice the height of that bounce:18 feet, ( 23 ).18 feet, ( 23 ).( 23 ).18 feet, ( 23 )3 .18 feet , ( 23 )4 .18 feet , . . . .The total distance traveled is the sum of the bounce–distances:total distance = 18 + ( 23 ).18 + ( 23 ).( 23 ).18 + ( 23 )3 .18 + ( 23 )4 .18+ . . .= 18 { 1 + 23 + ( 23 )2 + ( 23 )3 + ( 23 )4 + . . . }At the completion of the first bounce the ball has traveled 18 feet. After the second bounce, it hastravelled 30 feet, a total of 38 feet after the third bounce, 43 13 feet after the fourth, and so on. With acalculator and some patience, we see that after the 20th bounce the ball has traveled a total ofapproximately 53.996 feet, after the 30th bounce approximately 53.99994 feet, and after the 40thbounce approximately 53.9999989 feet.95...Height (feet)Bounce1234Fig. 1Bouncing Golf Ball(This is not the ball's path. The actual motion of the ball is straight up and down)10.2 Series Contemporary Calculus 2Practice 1:A tennis ball is thrown 10 feet straight upinto the air, and on each bounce it rebounds to 40%of its previous height. Represent the total distancetraveled by the ball as a sum, and find the total distancetraveled by the ball after the completion of its thirdbounce. (Fig. 2)Infinite SeriesThe infinite sums in the Example and Practice are called infinite series, and they are the objects we willstart to examine in this section.DefinitionAn infinite series is an expression of the forma1 + a2 + a3 + a4 + . . . or ∑k=1∞ ak .The numbers a1, a2, a3, a4, . . . are called the terms of the series. (Fig. 3)Example 2: Represent the following series using the sigma notation. (a) 1 + 1/3 + 1/9 + 1/27 + . . . ,(b) –1 + 1/2 – 1/3 + 1/4 – 1/5 + . . . , (c) 18( 2/3 + 4/9 + 8/27 + 16/81 + . . . )(d) 0.777 ... = 7/10 + 7 /100 + 7/1000 + ... , and (e) 0.222...Solution: (a) 1 + 1/3 + 1/9 + 1/27 + . . . = ∑k=0∞ ( 13 ) k or ∑k=1∞ ( 13 ) k–1(b) –1 + 1/2 – 1/3 + 1/4 – 1/5 + . . . = ∑k=1∞ (–1) k 1k (c) 18 ∑k=1∞ ( 23 ) k(d) 0.777 ... = 7/10 + 7 /100 + 7/1000 + ... = ∑k=1∞ 710k (e) ∑k=1∞ 210k Practice 2: Represent the following series using the sigma notation. (a) 1 + 2 + 3 + 4 + . . . ,(b) –1 + 1 – 1 + 1 – . . . (c) 2 + 1 + 1/2 + 1/4 + . . .(d) 1/2 + 1/4 + 1/6 + 1/8 + 1/10 +. . . (e) 0.111...104...Height (feet)Bounce1234Fig. 2Bouncing Tennis Ball(The actual motion of the ball is straight up and down)10.2 Series Contemporary Calculus 3In order to determine if the infinite series adds up to a finite value, we examine the sums as more and moreterms are added.DefinitionThe partial sums sn of the infinite series ∑k=1∞ ak are the numberss1 = a1,s2 = a1 + a2,s3 = a1 + a2 + a3 ,. . .In general, sn = a1 + a2 + a3 + . . . + an = ∑k=1n ak or , recursively, as sn = sn–1 + an .The partial sums form the sequence of partial sums { sn } .Example 3: Calculate the first 4 partial sums for the following series.(a) 1 + 1/2 + 1/4 + 1/8 + 1/16 + . . . , (b) ∑k=1∞ (–1) k , and (c) ∑n=1∞ 1n .Solution: (a) s1 = 1, s2 = 1 + 1/2 = 3/2, s3 = 1 + 1/2 + 1/4 = 7/4, s4 = 1 + 1/2 + 1/4 + 1/8 = 15/8It is usually easier to use the recursive version of sn :s3 = s2 + a3 = 3/2 + 1/4 = 7/4 and s4 = s3 + a4 = 7/4 + 1/8 = 15/8.(b) s1 = (–1)1 = –1, s2 = s1 + a2 = –1 + (–1)2 = 0, s3 = s2 + a3 = 0 + (–1)3 = –1, s4 = 0.(c) s1 = 1, s2 = 3/2, s3 = 11/6, s4 = 25/12 .Practice 3: Calculate the first 4 partial sums for the following series.(a) 1 – 1/2 + 1/4 – 1/8 + 1/16 – . . . , (b) ∑k=1∞ ( 13 ) k , and (c) ∑n=2∞ (–1)nn .If we know the values of the partial sums sn , we can recover the values of the terms an used to build the sn.Example 4: Suppose s1 = 2.1 , s2 = 2.6 , s3 = 2.84 , and s4 = 2.87 are the first partial sums of ∑k=1∞ ak .Find the values of the first four terms of { an } .Solution: s1 = a1 so a1 = 2.1 . s2 = a1 + a2 so 2.6 = 2.1 + a2 and a2 = 0.5 .Similarly, s3 = a1 + a2 + a3 so 2.84 = 2.1 + 0.5 + a3 and a3 = 0.24. Finally, a4 = 0.03 .42315610.5nansn1.5–0.5a2a3a4 a5s4s5a1s1=s2a1=+a2s3a1=+a2+a3s6Fig. 310.2 Series Contemporary Calculus 4An alternate solution method starts with a1 = s1 and then uses the fact that sn = sn–1 + an soan = sn – sn–1 . Thena2 = s2 – s1 = 2.6 – 2.1 = 0.5 .a3 = s3 – s2 = 2.84 – 2.6 = 0.24, anda4 = s4 – s3 = 2.87 – 2.84 = 0.03 .Practice 4: Suppose s1 = 3.2 , s2 = 3.6 , s3 = 3.5 , s4 = 4, s99 = 7.3 , s100 = 7.6, …