10.3 Geometric and Harmonic Series Contemporary Calculus 110.3 GEOMETRIC AND HARMONIC SERIESThis section uses ideas from Section 10.2 about series and their convergence to investigate some specialtypes of series. Geometric series are very important and appear in a variety of applications. Much of theearly work in the 17th century with series focused on geometric series and generalized them. Many of theideas used later in this chapter originated with geometric series. It is easy to determine whether a geometricseries converges or diverges, and when one does converge, we can easily find its sum. The harmonic seriesis important as an example of a divergent series whose terms approach zero. A final type of series, called"telescoping," is discussed briefly. Telescoping series are relatively uncommon, but their partial sumsexhibit a particularly nice pattern.Geometric Series: ∑k=0∞∞∞∞ C.r k = C + C.r + C.r2 + C.r3 + . . .Example 1: Bouncing Ball: A "super ball" is thrown 10 feetstraight up into the air. On each bounce, it rebounds tofour fifths of its previous height (Fig. 1) so the sequenceof heights is 10 feet, 8 feet, 32/5 feet, 128/25 feet, etc.(a) How far does the ball travel (up and down) duringits nth bounce. (b) Use a sum to represent the total distancetraveled by the ball.Solution: Since the ball travels up and down on each bounce,the distance traveled during each bounce is twice the heightof the ball on that bounce so d1 = 2(10 feet) = 20 feet,d2 = 16 feet, d3 = 64/5 feet, and, in general, dn = 45 .dn–1 .Looking at these values in another way,d1 = 20 , d2 = 45 .(20) , d3 = 45 d2 = 45 .45 .20 = ( 45 )2 (20) , d4 = 45 .d3 = 45 .( ( 45 )2 .(20) ) = ( 45 )3.(20) , and, in general, dn = ( 45 )n–1 .(20) .In theory, the ball bounces up and down forever, and the total distance traveled by the ball is the sum ofthe distances traveled during each bounce (an up and down flight):(first bounce) + (second bounce) + (third bounce) + (forth bounce) + . . .= 20 + 45 (20) + ( 45 )2 (20) + ( 45 )3 (20) + . . . )= 20.( 1 + 45 + ( 45 )2 + ( 45 )3 + . . . ) = 20. ∑k=0∞ ( 45 ) k .104...Height (feet)Bounce1234Fig. 1Bouncing "Super Ball"(The actual motion of the ball is straight up and down)8610.3 Geometric and Harmonic Series Contemporary Calculus 2Practice 1:Cake: Three calculus students want to share a small square cake equally, but they go about itin a rather strange way. First they cut the cake into 4 equal square pieces, each person takes one square,and one square is left (Fig. 2). Then they cut the leftover piece into 4 equal square pieces, each persontakes one square and one square is left. And they keep repeating this process. (a) What fraction of thetotal cake does each person "eventually" get? (b) Represent the amount of cake each person gets as ageometric series: (amount of first piece) + (amount of second piece) + . . .Each series in the previous Example and Practice problems is a Geometric series, a series in which each term is afixed multiple of the previous term. Geometric series have the form∑k=0∞ C.r k = C + C.r + C.r2 + C.r3 + . . . = C. ∑k=0∞ r kwith C ≠ 0 and r ≠ 0 representing fixed numbers. Each term in the series is r times the previous term.Geometric series are among the most common and easiest series we will encounter. A simple test determineswhether a geometric series converges, and we can even determine the "sum" of the geometric series.Geometric Series TheoremThe geometric series ∑k=0∞ r k = 1 + r + r2 + r3 + ...  converges to 11 – rif | r | < 1 diverges if | r | ≥ 1 Proof: If | r | ≥≥≥≥ 1, then | rk | approaches 1 or +∞ as k becomes arbitrarily large, so the terms ak = rk ofthe geometric series do not approach 0. Therefore, by the nth term test for divergence, the series diverges.ABCleftoverABCleftoverABCleftover. . .CakeThe cake is cut into four piecesA, B, and C each take a piece and one is leftoverThe leftover piece is cut into four piecesA, B, and C each take a piece and one is leftoverThe leftover piece is cut into four piecesFig. 2: Sharing a cake among three students10.3 Geometric and Harmonic Series Contemporary Calculus 3If | r | < 1, then the terms ak = rk of the geometric series approach 0 so the series may or maynot converge, and we need to examine the limit of the the partial sums sn = 1 + r + r2 + r3 + . . . +rn of the series. For a geometric series, a clever insight allows us to calculate those partial sums:(1 – r).sn = (1 – r).(1 + r + r2 + r3 + . . . + rn)= 1.(1 + r + r2 + r3 + . . . + rn) – r.(1 + r + r2 + r3 + . . . + rn)= (1 + r + r2 + r3 + . . . + rn) – ( r + r2 + r3 + r4 + . . . + rn + rn+1 )= 1 – rn+1 .Since | r | < 1 we know r ≠ 1 so we can divide the previous result by 1 – r to getsn = 1 + r + r2 + r3 + . . . + rn = 1 – rn+11 – r = 11 – r – rn+11 – r .This formula for the nth partial sum of a geometric series is sometimes useful, but now we are interestedin the limit of sn as n approaches infinity. Since | r | < 1, rn+1 approaches 0 as n approachesinfinity , we can conclude that the partial sums sn = 11 – r – rn+11 – r approach 11 – r (as "n→∞") .The geometric series ∑k=0∞ r k converges to the value 11 – r when –1 < r < 1 .Finally, ∑k=0∞ C.r k = C. ∑k=0∞ r k so we can easily determine whether or not ∑k=0∞ C.r kconverges and to what number.Example 2: How far did the ball in Example 1 travel?Solution: The distance traveled, 20( 1 + 45 + ( 45 )2 + ( 45 )3 + . . . ) , is a geometric series with C =20 and r = 4/5. Since | r | < 1, the series 1 + 45 + ( 45 )2 + ( 45 )3 + . . . converges to11 – r = 11 – 4/5 = 5, so the total distance traveled is20( 1 + 45 + ( 45 )2 + ( 45 )3 + . . . ) = 20( 5 ) = 100 feet.Repeating decimal numbers are really geometric series in disguise, and we can use the Geometric SeriesTheorem to represent the exact value of the sum as a fraction.Example 3: Represent the repeating …