#set document(title: "9.4 Infinite Geometric Series", author: "OpenStax / XYZ Homework") #set page(width: 8.5in, height: auto, margin: 1in) #import "@preview/cetz:0.5.2" #set text(font: ("STIX Two Text", "Libertinus Serif", "New Computer Modern"), size: 10.5pt, lang: "en") #show math.equation: set text(font: ("STIX Two Math", "New Computer Modern Math")) #set par(justify: true, leading: 0.62em, spacing: 0.9em) #set enum(spacing: 1.1em) // room between list items so tall inline fractions don't collide #set list(spacing: 1.1em) #set table(stroke: 0.5pt + rgb("#c7ccd3")) #let BLUE = rgb("#183B6F") // brand navy — section bars + example/solution labels (white on navy 11.09:1) #let ORANGE = rgb("#A94509") // brand primary-700 — AA-safe deep orange for TEXT (5.93:1 on white; raw brand #F37021 is 2.94:1 and must never carry text) #let RED = rgb("#DC2626") // brand error-600 #let GREEN = rgb("#059669") // brand success-600 (decoration only; small green text uses green-text #007942) #show heading.where(level: 1): it => block(width: 100%, above: 0pt, below: 16pt, fill: gradient.linear(BLUE, rgb("#2C5AA0")), inset: (x: 14pt, y: 12pt), radius: 3pt, text(fill: white, weight: "bold", size: 19pt, it.body)) #show heading.where(level: 2): it => block(width: 100%, above: 18pt, below: 10pt, fill: BLUE, inset: (x: 10pt, y: 6pt), radius: 2pt, text(fill: white, weight: "bold", size: 12pt, it.body)) #show heading.where(level: 3): it => text(fill: ORANGE, weight: "bold", size: 12.5pt, it.body) #show heading.where(level: 4): it => text(fill: BLUE, weight: "bold", size: 10.5pt, it.body) #let examplebox(label, title, body) = block(width: 100%, breakable: true, fill: rgb("#EFF1F5"), stroke: 0.5pt + rgb("#CFDDF0"), radius: 4pt, inset: 10pt, above: 12pt, below: 12pt)[ #block(below: 6pt)[#box(fill: BLUE, inset: (x: 6pt, y: 2pt), radius: 2pt, text(fill: white, weight: "bold", size: 8.5pt, label)) #h(0.4em) #strong[#title]] #body] // rail = decorative left rule (raw brand token); labelcolor = AA-safe label text shade #let notebox(label, rail, labelcolor, tint, body) = block(width: 100%, breakable: true, fill: tint, stroke: (left: 3pt + rail), inset: (left: 10pt, rest: 8pt), radius: (right: 4pt), above: 11pt, below: 11pt)[ #text(fill: labelcolor, weight: "bold", size: 7.5pt, tracking: 0.5pt)[#upper(label)] #linebreak() #body] #let solutionbox(body) = block(above: 4pt, below: 8pt)[ #text(fill: BLUE, weight: "bold", size: 8.5pt)[Solution] #linebreak() #body] #let figph(msg) = block(width: 100%, height: 60pt, fill: rgb("#f6f7f9"), stroke: (paint: rgb("#c7ccd3"), dash: "dashed"), radius: 4pt, inset: 10pt)[ #align(center + horizon, text(fill: rgb("#889"), style: "italic", size: 9pt, msg))] // Standardize inlined figure sizes: measure the natural CeTZ canvas, then scale to a // consistent envelope (aspect-aware; see build_typst.py FIG_* constants). Unlike the // print preamble, dimensions are FLOORED: in an editor a user can trim a figure to a // degenerate 1-D shape (a bare line), and w/h or tw/w would then divide by zero. #let _STD_W = 3.5 #let _WIDE_W = 5.6 #let _MAX_H = 3.4 #let _ASPECT_WIDE = 2.2 #let _UPSCALE_MAX = 1.15 #let stdfig(body) = context { let m = measure(body) let w = calc.max(m.width / 1in, 0.01) let h = calc.max(m.height / 1in, 0.01) let tw = if w / h > _ASPECT_WIDE { _WIDE_W } else { _STD_W } let s = calc.min(tw / w, _MAX_H / h, _UPSCALE_MAX) align(center, box(scale(x: s * 100%, y: s * 100%, reflow: true, body))) } #show figure: set block(breakable: false) #set figure(gap: 8pt) #show figure.caption: set text(size: 8.5pt, fill: rgb("#555")) == 9.4#h(0.6em)Infinite Geometric Series === Summation Notation There is a convenient notation for representing a series. For example, consider the sum of the first fifteen terms of the sequence #math.equation(block: true, alt: "4 , 7 , 10 , ⋯ , 3 n plus 1 , ⋯")[$4 , 7 , 10 , ⋯ , 3 n + 1 , ⋯$] To find the terms of the series we replace #math.equation(block: false, alt: "n")[$n$] in the general term #math.equation(block: false, alt: "3 n plus 1")[$3 n + 1$] by the numbers 1 through 15. So, instead of writing out all the terms, we might express the sum as #math.equation(block: true, alt: "The sum, as n runs from 1 to 15, of 3 n plus 1")[$"The sum, as" " " n " " "runs from 1 to 15, of" " " 3 n + 1$] We use the Greek letter #math.equation(block: false, alt: "Σ")[$upright(Σ)$] (called "sigma") to stand for "the sum," and we show the first and last values of #math.equation(block: false, alt: "n")[$n$] below and above the summation symbol #math.equation(block: false, alt: "Σ")[$upright(Σ)$], like this: #math.equation(block: true, alt: "∑ n equals 1153 n plus 1")[$limits(∑)^(15)_(n = 1) 3 n + 1$] Writing such an expression is sometimes called "using #strong[sigma notation]." The letter #math.equation(block: false, alt: "n")[$n$] is called the #strong[index of summation]; it is like a variable because it represents numerical values. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Any letter can be used for the index of summation; #math.equation(block: false, alt: "i , j ,")[$i , " " j ,$] and #math.equation(block: false, alt: "k")[$k$] are other common choices. Of course, the letter used for the index of summation does not affect the sum. ] #examplebox("Example 1")[][ Use sigma notation to represent the sum of the first 20 terms of the sequence #math.equation(block: true, alt: "minus 1 , 2 , 7 , ⋯ , k squared minus 2 , ⋯")[$− 1 , 2 , 7 , ⋯ , k^(2) − 2 , ⋯$] #solutionbox[ The general term of the sequence is #math.equation(block: false, alt: "k squared minus 2")[$k^(2) − 2$], and the first term is #math.equation(block: false, alt: "minus 1")[$− 1$], which we find by letting #math.equation(block: false, alt: "k equals 1")[$k = 1$] in the formula for the general term. Thus, #math.equation(block: true, alt: "∑ k equals 120 open parenthesis k squared minus 2 close parenthesis")[$limits(∑)^(20)_(k = 1) ( k^(2) − 2 )$] ] ] Use sigma notation to represent the sum of the first 20 terms of the sequence #math.equation(block: true, alt: "5 , 8 , 11 , and so on , 3 k plus 2 , and so on")[$5 , 8 , 11 , … , 3 k + 2 , …$] #math.equation(block: true, alt: "∑ k equals 1 b a sub k equals 5 plus 8 plus 11 plus ⋯ plus open parenthesis 3 k plus 2 close parenthesis plus ⋯")[$limits(∑)^(b)_(k = 1) a_(k) = 5 + 8 + 11 + ⋯ + ( 3 k + 2 ) + ⋯$] where #math.equation(block: false, alt: "b equals")[$b =$]\_\_\_\_\_ and #math.equation(block: false, alt: "a sub k equals")[$a_(k) =$]\_\_\_\_\_ #solutionbox[ #math.equation(block: true, alt: "∑ k equals 120 open parenthesis 3 k plus 2 close parenthesis")[$display(limits(∑)^(20)_(k = 1) ( 3 k + 2 ))$] ] Use sigma notation to represent the sum of the first 20 terms of the sequence #math.equation(block: true, alt: "5 , 8 , 11 , and so on , 3 k plus 2 , and so on")[$5 , 8 , 11 , … , 3 k + 2 , …$] #solutionbox[ #math.equation(block: true, alt: "∑ k equals 120 open parenthesis 3 k plus 2 close parenthesis")[$display(limits(∑)^(20)_(k = 1) ( 3 k + 2 ))$] ] What does the notation #math.equation(block: false, alt: "∑ j equals 18 the fraction 1 over j")[$display(limits(∑)^(8)_(j = 1) frac(1, j))$] mean? \_\_\_\_\_ #solutionbox[ Add the reciprocals of the numbers from 1 to 8. ] What does the notation #math.equation(block: false, alt: "∑ j equals 18 the fraction 1 over j")[$display(limits(∑)^(8)_(j = 1) frac(1, j))$] mean? + List 8 terms of the sequence #math.equation(block: false, alt: "a sub j equals the fraction 1 over j")[$a_(j) = display(frac(1, j))$]. + Add the numbers from 1 to 8 and take the reciprocal of the sum. + Add the reciprocals of the numbers from 1 to 8. + Add the first and eighth term of the sequence described. The #strong[expanded form] of a series written in sigma notation is obtained by writing out all the terms of the series. For example, #math.equation(block: true, alt: "∑ m equals 48 the fraction 3 over m equals the fraction 3 over 4 plus the fraction 3 over 5 plus the fraction 3 over 6 plus the fraction 3 over 7 plus the fraction 3 over 8")[$limits(∑)^(8)_(m = 4) display(frac(3, m)) = display(frac(3, 4)) + display(frac(3, 5)) + display(frac(3, 6)) + display(frac(3, 7)) + display(frac(3, 8))$] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Notice that the series above has five terms, which is one more than the difference of the upper and lower limits of summation, #math.equation(block: false, alt: "8 minus 4")[$8 − 4$]. ] Recall that the general term of an arithmetic series is a linear function of the index, and the general term of a geometric series is an exponential function. If we recognize a given series as one of these two types, we can use the formulas developed in the last section to evaluate the sum. #examplebox("Example 2")[][ Compute the value of each series. + #math.equation(block: false, alt: "∑ i equals 113 open parenthesis 90 minus 5 i close parenthesis")[$display(limits(∑)^(13)_(i = 1) ( 90 − 5 i ))$] + #math.equation(block: false, alt: "∑ k equals 092 to the power k")[$display(limits(∑)^(9)_(k = 0) 2^(k))$] #solutionbox[ + Because the general term #math.equation(block: false, alt: "90 minus 5 i")[$90 − 5 i$] is linear, this is an arithmetic series. By writing out the first few terms of the series, #math.equation(block: true, alt: "85 plus 80 plus 75 plus 70 plus ⋯")[$85 + 80 + 75 + 70 + ⋯$] we can verify that the first term of the series is #math.equation(block: false, alt: "a sub 1 equals 85")[$a_(1) = 85$] and the common difference is #math.equation(block: false, alt: "d equals minus 5")[$d = − 5$]. We also need to know the last term of the series, so we substitute #math.equation(block: false, alt: "n equals 13")[$n = 13$] in the general term to find #math.equation(block: false, alt: "a sub 13 equals 90 minus 5 open parenthesis 13 close parenthesis equals 25")[$a_(13) = 90 − 5 ( 13 ) = 25$] Thus, #math.equation(block: true, alt: "∑ i equals 113 open parenthesis 90 minus 5 i close parenthesis equals the fraction 13 over 2 open parenthesis 85 plus 25 close parenthesis equals 715")[$limits(∑)^(13)_(i = 1) ( 90 − 5 i ) = display(frac(13, 2)) ( 85 + 25 ) = 715$] + The general term of this series is exponential, so the series is geometric. By writing out a few terms of the series, #math.equation(block: true, alt: "1 plus 2 plus 4 plus 8 plus ⋯")[$1 + 2 + 4 + 8 + ⋯$] we confirm that the first term of the series is #math.equation(block: false, alt: "a sub 1 equals 1")[$a_(1) = 1$] and the common ratio is #math.equation(block: false, alt: "r equals 2")[$r = 2$]. The series has 10 terms, from #math.equation(block: false, alt: "k equals 0")[$k = 0$] to #math.equation(block: false, alt: "k equals 9")[$k = 9$], so #math.equation(block: false, alt: "n equals 10")[$n = 10$]. Finally, we substitute these values into the formula for geometric series to obtain #math.equation(block: true, alt: "∑ k equals 092 to the power k equals the fraction 1 open parenthesis 1 minus 2 to the power 10 close parenthesis over 1 minus 2 equals 1023")[$limits(∑)^(9)_(k = 0) 2^(k) = display(frac(1 ( 1 − 2^(10) ), 1 − 2)) = 1023$] ] ] Compute the value of each series. + #math.equation(block: false, alt: "∑ k equals 150 open parenthesis 3 k plus 2 close parenthesis equals")[$display(limits(∑)^(50)_(k = 1) ( 3 k + 2 ) =)$]\_\_\_\_\_ + #math.equation(block: false, alt: "∑ n equals 1810 to the power n equals")[$display(limits(∑)^(8)_(n = 1) 10^(n) =)$]\_\_\_\_\_ Do not use any commas. For example, use "10000" rather than "10,000". #solutionbox[ + 3925 + 111,111,110 ] Compute the value of each series. + #math.equation(block: false, alt: "∑ k equals 150 open parenthesis 3 k plus 2 close parenthesis")[$display(limits(∑)^(50)_(k = 1) ( 3 k + 2 ))$] + #math.equation(block: false, alt: "∑ n equals 1810 to the power n")[$display(limits(∑)^(8)_(n = 1) 10^(n))$] #solutionbox[ + 3925 + 111,111,110 ] If the series is not arithmetic or geometric, we don't have a formula for the sum, so we must compute the sum directly. #examplebox("Example 3")[][ Compute the value of each series. + #math.equation(block: false, alt: "∑ m equals 15 m squared")[$display(limits(∑)^(5)_(m = 1) m^(2))$] + #math.equation(block: false, alt: "∑ p equals 18005")[$display(limits(∑)^(800)_(p = 1) 5)$] #solutionbox[ + Because the general term, #math.equation(block: false, alt: "m squared")[$m^(2)$], is neither linear nor exponential, we know that the series is not arithmetic or geometric. However, we can expand the series and evaluate it directly. #math.equation(block: true, alt: "∑ m equals 15 m squared, equals 1 squared plus 2 squared plus 3 squared plus 4 squared plus 5 squared; equals 1 plus 4 plus 9 plus 16 plus 25 equals 55")[$display(limits(∑)^(5)_(m = 1) m^(2)) & = 1^(2) + 2^(2) + 3^(2) + 4^(2) + 5^(2) \ & = 1 + 4 + 9 + 16 + 25 = 55$] + The general term is the number 5. Because the index #math.equation(block: false, alt: "p")[$p$] runs from 1 to 800, we are adding 800 terms, each of which is 5. Thus #math.equation(block: true, alt: "∑ p equals 18005, equals 5 plus 5 plus 5 plus ⋯ plus 5 plus 5; equals 800 open parenthesis 5 close parenthesis equals 4000")[$display(limits(∑)^(800)_(p = 1) 5) & = 5 + 5 + 5 + ⋯ + 5 + 5 \ & = 800 ( 5 ) = 4000$] ] ] Compute the value of each series. + #math.equation(block: false, alt: "∑ k equals 020 the fraction 1 over 3 equals")[$display(limits(∑)^(20)_(k = 0) display(frac(1, 3)) =)$]\_\_\_\_\_ + #math.equation(block: false, alt: "∑ k equals 03 the fraction k over k plus 1 equals")[$display(limits(∑)^(3)_(k = 0) display(frac(k, k + 1)) =)$]\_\_\_\_\_ #solutionbox[ + 7 + #math.equation(block: false, alt: "the fraction 23 over 12")[$display(frac(23, 12))$] ] Compute the value of each series. + #math.equation(block: false, alt: "∑ k equals 020 the fraction 1 over 3")[$display(limits(∑)^(20)_(k = 0) display(frac(1, 3)))$] + #math.equation(block: false, alt: "∑ k equals 03 the fraction k over k plus 1")[$display(limits(∑)^(3)_(k = 0) display(frac(k, k + 1)))$] #solutionbox[ + #math.equation(block: false, alt: "7")[$7$] + #math.equation(block: false, alt: "the fraction 23 over 12")[$display(frac(23, 12))$] ] Which of these series is neither arithmetic nor geometric? \_\_\_\_\_ #solutionbox[ #math.equation(block: false, alt: "∑ n equals 212 open parenthesis n cubed plus 2 close parenthesis")[$display(limits(∑)^(12)_(n = 2) ( n^(3) + 2 ))$] is neither arithmetic nor geometric. (The first series is arithmetic; the second and fourth series are both geometric. ] Which of these series is neither arithmetic nor geometric? \` + #math.equation(block: false, alt: "∑ k equals 515 open parenthesis the fraction 1 over 2 k minus 4 close parenthesis")[$display(limits(∑)^(15)_(k = 5) ( frac(1, 2) k − 4 ))$] + #math.equation(block: false, alt: "∑ m equals 1500.8 to the power m")[$display(limits(∑)^(50)_(m = 1) 0.8^(m))$] + #math.equation(block: false, alt: "∑ n equals 212 open parenthesis n cubed plus 2 close parenthesis")[$display(limits(∑)^(12)_(n = 2) ( n^(3) + 2 ))$] + #math.equation(block: false, alt: "∑ p equals 155 open parenthesis 1.5 close parenthesis to the power 2 p")[$display(limits(∑)^(5)_(p = 1) 5 ( 1.5 )^(2 p))$] === Infinite Series A series with infinitely many terms is called an #strong[infinite series]. Is it possible to add infinitely many terms and arrive at a finite sum? In some cases, if the terms added are small enough, the answer is yes. The question above, answered by watching partial sums. The blue curve passes through the nth partial sum Sₙ of the geometric series r + r² + r³ + ⋯ at every whole number n; the dashed line is the value a/(1 − r) that the formula proposes for the total. At r = 0.5 this is exactly the section's series ½ + ¼ + ⅛ + ⋯: the curve climbs toward the dashed line at height 1 but never crosses it — each new term is half the remaining gap. For any r below 1 on the slider the same thing happens at height r/(1 − r). Now slide r above 1: the curve blows up — the terms themselves grow, so the partial sums climb without bound and the series has no sum. The formula agrees, in its own way: for r above 1 it returns a negative number, and the dashed line falls below the x-axis (visibly at −3.5 when r = 1.4; even farther down, off the bottom of the frame, at r = 1.1) — a sum of positive terms cannot be negative, the algebra's own warning that the formula only means something when |r| \< 1.{"functions":\[{"color":"\#6a6a6a","expression":"r/(1 - r)","expression2":"","id":"sum-line","inequality":"lt","lineDash":\[8,5\],"lineWidth":1.5,"mode":"cartesian","tMax":10,"tMin":-10,"variable":"x","visible":true},{"color":"\#1f4e79","domain":\[0,12\],"expression":"r\*(1 - r^x)/(1 - r)","expression2":"","id":"partial-sums","inequality":"lt","lineDash":\[\],"lineWidth":2.5,"mode":"cartesian","tMax":12,"tMin":0,"variable":"x","visible":true}\],"grid":{"majorSpacing":1,"showAxes":true,"showLabels":true},"parameters":\[{"id":"p-r","label":"Common ratio r","max":1.4,"min":0.2,"name":"r","step":0.3,"value":0.5}\],"title":"Partial sums of r + r\\u00b2 + r\\u00b3 + \\u22ef and the value a/(1 \\u2212 r)","version":1,"viewport":{"centerX":5,"centerY":2,"scale":40}}Consider the infinite geometric series #math.equation(block: true, alt: "the fraction 1 over 2 plus the fraction 1 over 4 plus the fraction 1 over 8 plus the fraction 1 over 16 plus ⋯")[$display(frac(1, 2)) + display(frac(1, 4)) + display(frac(1, 8)) + display(frac(1, 16)) + ⋯$] The #math.equation(block: false, alt: "n to the power th")[$n^("th")$] #strong[partial sum] of the series is the sum of its first #math.equation(block: false, alt: "n")[$n$] terms, and is denoted by #math.equation(block: false, alt: "S sub n")[$S_(n)$]. Thus, #math.equation(block: true, alt: "S sub 1, equals the fraction 1 over 2; S sub 2, equals the fraction 1 over 2 plus the fraction 1 over 4 equals the fraction 3 over 4; S sub 3, equals the fraction 1 over 2 plus the fraction 1 over 4 plus the fraction 1 over 8 equals the fraction 7 over 8; S sub 4, equals the fraction 1 over 2 plus the fraction 1 over 4 plus the fraction 1 over 8 plus the fraction 1 over 16 equals the fraction 15 over 16")[$S_(1) & = display(frac(1, 2)) \ S_(2) & = display(frac(1, 2)) + display(frac(1, 4)) = display(frac(3, 4)) \ S_(3) & = display(frac(1, 2)) + display(frac(1, 4)) + display(frac(1, 8)) = display(frac(7, 8)) \ S_(4) & = display(frac(1, 2)) + display(frac(1, 4)) + display(frac(1, 8)) + display(frac(1, 16)) = display(frac(15, 16))$] Notice that for each partial sum, the new term added gets smaller and smaller: #math.equation(block: false, alt: "the fraction 1 over 2")[$display(frac(1, 2))$], #math.equation(block: false, alt: "the fraction 1 over 4")[$display(frac(1, 4))$], #math.equation(block: false, alt: "the fraction 1 over 8")[$display(frac(1, 8))$], #math.equation(block: false, alt: "the fraction 1 over 16 ,")[$display(frac(1, 16)) ,$] and so on. You can also see that as #math.equation(block: false, alt: "n")[$n$] increases and we add more and more terms of the series, the partial sums are getting closer to 1. In fact, as #math.equation(block: false, alt: "n")[$n$] becomes very large, #math.equation(block: false, alt: "S sub n")[$S_(n)$] gets very close to 1. It seems reasonable that the sum of all the terms of the series is 1. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Will the partial sums ever be greater than 1? No, because each new term #math.equation(block: false, alt: "a sub n")[$a_(n)$] added is half the difference between #math.equation(block: false, alt: "S sub n")[$S_(n)$] and 1. ] We can make this conjecture more plausible by examining the formula for the #math.equation(block: false, alt: "n to the power th")[$n^("th")$] partial sum of a geometric series, #math.equation(block: true, alt: "S sub n, equals the fraction a sub n plus 1 minus a sub 1 over r minus 1; equals the fraction a r to the power n minus a over r minus 1 equals the fraction a minus a r to the power n over 1 minus r, open parenthesis r not equal to 1 close parenthesis")[$S_(n) & = display(frac(a_(n + 1) − a_(1), r − 1)) \ & = display(frac(a r^(n) − a, r − 1)) = display(frac(a − a r^(n), 1 − r)) & & ( r ≠ 1 )$] Look at the second term of the numerator, #math.equation(block: false, alt: "a r to the power n")[$a r^(n)$]. This term is the only part of the formula that depends on #math.equation(block: false, alt: "n")[$n$]. What happens to #math.equation(block: false, alt: "a r to the power n")[$a r^(n)$] as #math.equation(block: false, alt: "n")[$n$] increases? Consider two examples: - If #math.equation(block: false, alt: "r equals the fraction 1 over 2")[$r = display(frac(1, 2))$], then #math.equation(block: true, alt: "r squared equals open parenthesis the fraction 1 over 2 close parenthesis squared equals the fraction 1 over 4 , r cubed equals open parenthesis the fraction 1 over 2 close parenthesis cubed equals the fraction 1 over 8 , r to the power 4 equals open parenthesis the fraction 1 over 2 close parenthesis to the power 4 equals the fraction 1 over 16 ,")[$r^(2) = attach(( display(frac(1, 2)) ), t: 2) = display(frac(1, 4)) , #h(2em) r^(3) = attach(( display(frac(1, 2)) ), t: 3) = display(frac(1, 8)) , #h(2em) r^(4) = attach(( display(frac(1, 2)) ), t: 4) = display(frac(1, 16)) ,$] and so on, with #math.equation(block: false, alt: "open parenthesis the fraction 1 over 2 close parenthesis to the power n")[$attach(( display(frac(1, 2)) ), t: n)$] becoming smaller and smaller for larger values of #math.equation(block: false, alt: "n")[$n$]. Each time we multiply by another factor of #math.equation(block: false, alt: "r equals the fraction 1 over 2")[$r = display(frac(1, 2))$], the product gets smaller, because #math.equation(block: false, alt: "the fraction 1 over 2 less than 1")[$display(frac(1, 2)) < 1$]. - On the other hand, if #math.equation(block: false, alt: "r greater than 1")[$r > 1$], multiplying by #math.equation(block: false, alt: "r")[$r$] makes the product larger. If #math.equation(block: false, alt: "r equals the fraction 3 over 2")[$r = display(frac(3, 2))$] for example, then #math.equation(block: true, alt: "r squared equals open parenthesis the fraction 3 over 2 close parenthesis squared equals the fraction 9 over 4 , r cubed equals open parenthesis the fraction 3 over 2 close parenthesis cubed equals the fraction 27 over 8 , r to the power 4 equals open parenthesis the fraction 3 over 2 close parenthesis to the power 4 equals the fraction 81 over 16 ,")[$r^(2) = attach(( display(frac(3, 2)) ), t: 2) = display(frac(9, 4)) , #h(2em) r^(3) = attach(( display(frac(3, 2)) ), t: 3) = display(frac(27, 8)) , #h(2em) r^(4) = attach(( display(frac(3, 2)) ), t: 4) = display(frac(81, 16)) ,$] and so on. In this case the powers of #math.equation(block: false, alt: "r")[$r$] are increasing. In general, we have the following result. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Increasing #math.equation(block: false, alt: "r to the power n")[$r^(n)$]] + If #math.equation(block: false, alt: "0 less than vertical bar r vertical bar less than 1")[$0 < | r | < 1$], then #math.equation(block: false, alt: "r to the power n")[$r^(n)$] gets closer to zero as #math.equation(block: false, alt: "n")[$n$] increases. + If #math.equation(block: false, alt: "vertical bar r vertical bar greater than 1")[$| r | > 1$], then #math.equation(block: false, alt: "r to the power n")[$r^(n)$] does not approach a finite number as #math.equation(block: false, alt: "n")[$n$] increases. ] Now let's return to the formula for #math.equation(block: false, alt: "S sub n")[$S_(n)$] and write it in the form #math.equation(block: true, alt: "S sub n equals the fraction a over 1 minus r open parenthesis 1 minus r to the power n close parenthesis")[$S_(n) = display(frac(a, 1 − r)) ( 1 − r^(n) )$] where we have factored #math.equation(block: false, alt: "a")[$a$] from the numerator. There are two cases to consider. If #math.equation(block: false, alt: "vertical bar r vertical bar less than 1")[$| r | < 1$], then #math.equation(block: false, alt: "r to the power n")[$r^(n)$] approaches 0, and the factor #math.equation(block: false, alt: "1 minus r to the power n")[$1 − r^(n)$] gets closer and closer to 1 as #math.equation(block: false, alt: "n")[$n$] grows larger. Consequently, if we compute #math.equation(block: false, alt: "S sub n")[$S_(n)$] for larger and larger values of #math.equation(block: false, alt: "n")[$n$], the sum approaches the value #math.equation(block: true, alt: "the fraction a over 1 minus r")[$display(frac(a, 1 − r))$] This analysis motivates us to define the sum of an infinite geometric series as follows. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Sum of an Infinite Geometric Series] The sum of an infinite geometric series #math.equation(block: false, alt: "∑ k equals 0 ∞ a r to the power k minus 1")[$" " " " display(limits(∑)^(upright(∞))_(k = 0) a r^(k − 1) " " " ")$] is #math.equation(block: true, alt: "S sub ∞ equals the fraction a over 1 minus r if minus 1 less than r less than 1")[$S_(upright(∞)) = display(frac(a, 1 − r)) " " " " " " " " " " " " "if" " " " " " " " " − 1 < r < 1$] ] In the second case, if #math.equation(block: false, alt: "vertical bar r vertical bar greater than 1")[$| r | > 1$], as in the infinite series #math.equation(block: true, alt: "3 plus 6 plus 12 plus ⋯")[$3 + 6 + 12 + ⋯$] where #math.equation(block: false, alt: "r equals 2")[$r = 2$], the terms become larger as #math.equation(block: false, alt: "n")[$n$] increases, and the sum of the series is not a finite number. In this case the series does not have a sum. Under what circumstances does the series #math.equation(block: false, alt: "∑ k equals 0 ∞ a r to the power k")[$display(limits(∑)^(upright(∞))_(k = 0) a r^(k))$] have a finite sum? \_\_\_\_\_ #solutionbox[ If #math.equation(block: false, alt: "vertical bar r vertical bar less than 1")[$| r | < 1$] ] Under what circumstances does the series #math.equation(block: false, alt: "∑ k equals 0 ∞ a r to the power k")[$display(limits(∑)^(upright(∞))_(k = 0) a r^(k))$] have a finite sum? + If #math.equation(block: false, alt: "a less than 1")[$a < 1$] + If #math.equation(block: false, alt: "r less than 1")[$r < 1$] + If #math.equation(block: false, alt: "vertical bar r vertical bar less than 1")[$| r | < 1$] + If #math.equation(block: false, alt: "a r less than r")[$a r < r$] #examplebox("Example 4")[][ Make a table showing the first five partial sums of each series. Then use the formula to find the sum, if it exists. + #math.equation(block: false, alt: "∑ j equals 0 ∞ 30 open parenthesis 0.8 close parenthesis to the power j")[$" " " " display(limits(∑)^(upright(∞))_(j = 0) 30 ( 0.8 )^(j) " " " ")$] + #math.equation(block: false, alt: "∑ m equals 0 ∞ 3 open parenthesis the fraction 4 over 3 close parenthesis to the power m")[$" " " " display(limits(∑)^(upright(∞))_(m = 0) 3 attach(( display(frac(4, 3)) ), t: m) " " " ")$] #solutionbox[ + First, we evaluate the general term #math.equation(block: false, alt: "30 open parenthesis 0.8 close parenthesis to the power j")[$30 ( 0.8 )^(j)$] for #math.equation(block: false, alt: "j equals 1 , 2 , ⋯ , 5")[$j = 1 , 2 , ⋯ , 5$] and compute the partial sums. For example, #math.equation(block: true, alt: "S sub 1, equals 30 open parenthesis 0.8 close parenthesis to the power 1 equals 24; S sub 2, equals 30 open parenthesis 0.8 close parenthesis to the power 1 plus 30 open parenthesis 0.8 close parenthesis squared equals 43.2")[$S_(1) & = 30 ( 0.8 )^(1) = 24 \ S_(2) & = 30 ( 0.8 )^(1) + 30 ( 0.8 )^(2) = 43.2$] and so on. The results, rounded to hundredths, are shown in the table.#figure(table( columns: 6, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "n")[$n$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "5")[$5$]]), [#math.equation(block: false, alt: "S sub n")[$S_(n)$]], [#math.equation(block: false, alt: "24")[$24$]], [#math.equation(block: false, alt: "43.2")[$43.2$]], [#math.equation(block: false, alt: "58.56")[$58.56$]], [#math.equation(block: false, alt: "70.85")[$70.85$]], [#math.equation(block: false, alt: "80.68")[$80.68$]], )) #linebreak() This is an infinite geometric series with #math.equation(block: false, alt: "a equals 30")[$a = 30$] and #math.equation(block: false, alt: "r equals 0.8")[$r = 0.8$]. The series has a sum because #math.equation(block: false, alt: "vertical bar r vertical bar less than 1")[$| r | < 1$]. Thus, #math.equation(block: true, alt: "S sub ∞ equals the fraction a over 1 minus r equals the fraction 30 over 1 minus 0.8 equals 150")[$S_(upright(∞)) = display(frac(a, 1 − r)) = display(frac(30, 1 − 0.8)) = 150$] + We evaluate the general term #math.equation(block: false, alt: "3 open parenthesis the fraction 4 over 3 close parenthesis to the power m")[$3 attach(( display(frac(4, 3)) ), t: m)$] for #math.equation(block: false, alt: "m equals 1 , 2 , ⋯ , 5")[$m = 1 , 2 , ⋯ , 5$] and compute the partial sums. For example, #math.equation(block: true, alt: "S sub 1, equals 3 open parenthesis the fraction 4 over 3 close parenthesis to the power 1 equals 4; S sub 2, equals 3 open parenthesis the fraction 4 over 3 close parenthesis to the power 1 plus 3 open parenthesis the fraction 4 over 3 close parenthesis squared equals the fraction 28 over 3")[$S_(1) & = 3 attach(( display(frac(4, 3)) ), t: 1) = 4 \ S_(2) & = 3 attach(( display(frac(4, 3)) ), t: 1) + 3 attach(( display(frac(4, 3)) ), t: 2) = display(frac(28, 3))$] and so on. The results, rounded to hundredths, are shown in the table.#figure(table( columns: 6, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "n")[$n$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "5")[$5$]]), [#math.equation(block: false, alt: "S sub n")[$S_(n)$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "9.33")[$9.33$]], [#math.equation(block: false, alt: "16.44")[$16.44$]], [#math.equation(block: false, alt: "25.93")[$25.93$]], [#math.equation(block: false, alt: "38.57")[$38.57$]], )) #linebreak() This is a geometric series with #math.equation(block: false, alt: "a equals 3")[$a = 3$] and #math.equation(block: false, alt: "r equals the fraction 4 over 3")[$r = display(frac(4, 3))$]. The infinite series does not have a sum because #math.equation(block: false, alt: "vertical bar r vertical bar greater than 1")[$| r | > 1$]. ] ] Find the sum, if it exists. If the series does not have a sum, enter "no sum". + #math.equation(block: false, alt: "∑ j equals 0 ∞ 13 open parenthesis the fraction 7 over 6 close parenthesis to the power j equals")[$display(limits(∑)^(upright(∞))_(j = 0) 13 attach(( display(frac(7, 6)) ), t: j) =)$]\_\_\_\_\_ + #math.equation(block: false, alt: "∑ m equals 0 ∞ 5.9 open parenthesis 0.9 close parenthesis to the power m equals")[$display(limits(∑)^(upright(∞))_(m = 0) 5.9 ( 0.9 )^(m) =)$]\_\_\_\_\_ #solutionbox[ + No sum + 59 ] Find the sum, if it exists. + #math.equation(block: false, alt: "∑ j equals 0 ∞ 13 open parenthesis the fraction 7 over 6 close parenthesis to the power j")[$display(limits(∑)^(upright(∞))_(j = 0) 13 attach(( display(frac(7, 6)) ), t: j))$] + #math.equation(block: false, alt: "∑ m equals 0 ∞ 5.9 open parenthesis 0.9 close parenthesis to the power m")[$display(limits(∑)^(upright(∞))_(m = 0) 5.9 ( 0.9 )^(m))$] #solutionbox[ + No sum + 59 ] === Repeating Decimals An interesting application of geometric series involves repeating decimals. Recall that the decimal representation of a rational number either terminates, as does #math.equation(block: false, alt: "0.75 ,")[$0.75 ,$] or repeats a pattern of digits. For example, you probably recognize #math.equation(block: false, alt: "0.3333 bar")[$0.333 limits(3)^(―)$] as the decimal representation of #math.equation(block: false, alt: "the fraction 1 over 3")[$display(frac(1, 3))$]. It is easy to find the decimal form of a fraction: we just divide the denominator into the numerator. Is there a way to find the fractional form of a repeating decimal? Consider the repeating decimal #math.equation(block: true, alt: "0.212121 bar")[$0.2121 limits(21)^(―)$] We can write this number as an infinite geometric series: #math.equation(block: true, alt: "0.21 plus 0.0021 plus 0.000021 plus ⋯")[$0.21 + 0.0021 + 0.000021 + ⋯$] The first term of this series is #math.equation(block: false, alt: "0.21")[$0.21$], or #math.equation(block: false, alt: "the fraction 21 over 100")[$display(frac(21, 100))$], and its common ratio is #math.equation(block: false, alt: "r equals 0.01")[$r = 0.01$], or #math.equation(block: false, alt: "the fraction 1 over 100")[$display(frac(1, 100))$]. Because #math.equation(block: false, alt: "vertical bar r vertical bar less than 1")[$| r | < 1$], the series has a sum given by #math.equation(block: true, alt: "S sub ∞, equals the fraction a over 1 minus r equals the fraction the fraction 21 over 100 over 1 minus the fraction 1 over 100; equals the fraction the fraction 21 over 100 over the fraction 99 over 100 equals the fraction 21 over 99 equals the fraction 7 over 33")[$S_(upright(∞)) & = display(frac(a, 1 − r)) = display(frac(display(frac(21, 100)), 1 − display(frac(1, 100)))) \ & = display(frac(display(frac(21, 100)), display(frac(99, 100)))) = display(frac(21, 99)) = display(frac(7, 33))$] Thus, the decimal number #math.equation(block: false, alt: "0.212121 bar")[$0.2121 limits(21)^(―)$] is equal to the fraction #math.equation(block: false, alt: "the fraction 7 over 33")[$display(frac(7, 33))$]. #examplebox("Example 5")[][ Find a common fraction equivalent to #math.equation(block: false, alt: "0.37 bar")[$0.3 limits(7)^(―)$]. #solutionbox[ The decimal can be written as #math.equation(block: false, alt: "0.3 plus 0.07 bar")[$0.3 + 0.0 limits(7)^(―)$]. We will find a fraction equivalent to the repeating decimal #math.equation(block: false, alt: "0.07 bar")[$0.0 limits(7)^(―)$] and add that to #math.equation(block: false, alt: "0.3")[$0.3$], or #math.equation(block: false, alt: "the fraction 3 over 10")[$display(frac(3, 10))$]. We write #math.equation(block: false, alt: "0.07 bar")[$0.0 limits(7)^(―)$] as a series: #math.equation(block: true, alt: "0.07 bar equals the fraction 7 over 100 plus the fraction 7 over 1000 plus the fraction 7 over 10 , 000 plus ⋯")[$0.0 limits(7)^(―) = display(frac(7, 100)) + display(frac(7, 1000)) + display(frac(7, 10 "," 000)) + ⋯$] This is an infinite geometric series with first term #math.equation(block: false, alt: "the fraction 7 over 100")[$display(frac(7, 100))$] and common ratio #math.equation(block: false, alt: "the fraction 1 over 10")[$display(frac(1, 10))$]. The sum of the series is given by #math.equation(block: true, alt: "S sub ∞, equals the fraction a over 1 minus r equals the fraction the fraction 7 over 100 over 1 minus the fraction 1 over 10; equals the fraction the fraction 7 over 100 over the fraction 9 over 10 equals the fraction 7 over 100 times the fraction 9 over 10 equals the fraction 7 over 90")[$S_(upright(∞)) & = display(frac(a, 1 − r)) = display(frac(display(frac(7, 100)), 1 − display(frac(1, 10)))) \ & = display(frac(display(frac(7, 100)), display(frac(9, 10)))) = display(frac(7, 100)) ⋅ display(frac(9, 10)) = display(frac(7, 90))$] Finally, we add #math.equation(block: false, alt: "the fraction 7 over 90")[$display(frac(7, 90))$] and #math.equation(block: false, alt: "the fraction 3 over 10")[$display(frac(3, 10))$] to find #math.equation(block: true, alt: "0.37 bar equals the fraction 3 over 10 plus the fraction 7 over 90 equals the fraction 34 over 90")[$0.3 limits(7)^(―) = display(frac(3, 10)) + display(frac(7, 90)) = display(frac(34, 90))$] ] ] Find a common fraction equivalent to #math.equation(block: false, alt: "0.8 bar")[$0. limits(8)^(―)$]: \_\_\_\_\_ #solutionbox[ #math.equation(block: true, alt: "the fraction 8 over 9")[$display(frac(8, 9))$] ] Find a common fraction equivalent to #math.equation(block: false, alt: "0.8 bar")[$0. limits(8)^(―)$]. #solutionbox[ #math.equation(block: true, alt: "the fraction 8 over 9")[$display(frac(8, 9))$] ] === Section Summary ==== Vocabulary - Sigma notation - Index of summation - Infinite series - Partial sum ==== CONCEPTS + We can use sigma notation to denote a series. + It is possible to add infinitely many terms and arrive at a finite sum if the terms are small enough. + The sum of an infinite geometric series #math.equation(block: false, alt: "∑ k equals 0 ∞ a r to the power k minus 1")[$" " " " display(limits(∑)^(upright(∞))_(k = 0) a r^(k − 1) " " " ")$] is #math.equation(block: true, alt: "S sub ∞ equals the fraction a over 1 minus r if minus 1 less than r less than 1")[$S_(upright(∞)) = display(frac(a, 1 − r)) " " " " " " " " " " " " "if" " " " " " " " " − 1 < r < 1$] + If #math.equation(block: false, alt: "vertical bar r vertical bar greater than 1")[$| r | > 1$] in an infinite geometric series, the series does not have a sum. ==== STUDY QUESTIONS + Explain how to use sigma notation to define a series. + What is an infinite series? + What is a partial sum of an infinite series? + State a formula for evaluating an infinite geometric series. Under what conditions is the formula valid? + A repeating decimal can be rewritten as what kind of series? ==== SKILLS Practice each skill in the Homework Problems listed. + Express sigma notation in expanded form: \#1-8 + Write a series in sigma notation: \#12-20 + Identify a series as arithmetic, geometric, or neither and evaluate: \#21-40 + Evaluate an infinite geometric series: \#41-48, \#57-60 + Write a repeating decimal as a common fraction: \#49-56 === Homework 9.4 For Problems 1–8, write the sum in expanded form. #math.equation(block: true, alt: "∑ i equals 14 i squared")[$display(limits(∑)^(4)_(i = 1) i^(2))$] #math.equation(block: true, alt: "1 squared plus 2 squared plus 3 squared plus 4 squared")[$1^(2) + 2^(2) + 3^(2) + 4^(2)$] #math.equation(block: true, alt: "∑ i equals 13 open parenthesis 3 i minus 2 close parenthesis")[$display(limits(∑)^(3)_(i = 1) ( 3 i − 2 ))$] #math.equation(block: true, alt: "∑ j equals 57 open parenthesis j minus 2 close parenthesis")[$display(limits(∑)^(7)_(j = 5) ( j − 2 ))$] #math.equation(block: true, alt: "3 plus 4 plus 5")[$3 + 4 + 5$] #math.equation(block: true, alt: "∑ j equals 26 open parenthesis j squared plus 1 close parenthesis")[$display(limits(∑)^(6)_(j = 2) ( j^(2) + 1 ))$] #math.equation(block: true, alt: "∑ k equals 14 k open parenthesis k plus 1 close parenthesis")[$display(limits(∑)^(4)_(k = 1) k ( k + 1 ))$] #math.equation(block: true, alt: "1 open parenthesis 2 close parenthesis plus 2 open parenthesis 3 close parenthesis plus 3 open parenthesis 4 close parenthesis plus 4 open parenthesis 5 close parenthesis")[$1 ( 2 ) + 2 ( 3 ) + 3 ( 4 ) + 4 ( 5 )$] #math.equation(block: true, alt: "∑ k equals 26 the fraction k over 2 open parenthesis k plus 1 close parenthesis")[$display(limits(∑)^(6)_(k = 2) display(frac(k, 2)) ( k + 1 ))$] #math.equation(block: true, alt: "∑ m equals 14 the fraction open parenthesis minus 1 close parenthesis to the power m over 2 to the power m")[$display(limits(∑)^(4)_(m = 1) display(frac(( − 1 )^(m), 2^(m))))$] #math.equation(block: true, alt: "the fraction minus 1 over 2 plus the fraction 1 over 2 squared minus the fraction 1 over 2 cubed plus the fraction 1 over 2 to the power 4")[$display(frac(− 1, 2)) + display(frac(1, 2^(2))) − display(frac(1, 2^(3))) + display(frac(1, 2^(4)))$] #math.equation(block: true, alt: "∑ m equals 35 the fraction open parenthesis minus 1 close parenthesis to the power m plus 1 over m minus 2")[$display(limits(∑)^(5)_(m = 3) display(frac(( − 1 )^(m + 1), m − 2)))$] For Problems 9–20, write the series using sigma notation. #math.equation(block: true, alt: "1 plus 3 plus 5 plus 7")[$1 + 3 + 5 + 7$] #math.equation(block: true, alt: "∑ k equals 14 open parenthesis 2 k minus 1 close parenthesis")[$display(limits(∑)^(4)_(k = 1) ( 2 k − 1 ))$] #math.equation(block: true, alt: "2 plus 4 plus 6 plus 8")[$2 + 4 + 6 + 8$] #math.equation(block: true, alt: "5 plus 5 cubed plus 5 to the power 5 plus 5 to the power 7")[$5 + 5^(3) + 5^(5) + 5^(7)$] #math.equation(block: true, alt: "∑ k equals 145 to the power 2 k minus 1")[$display(limits(∑)^(4)_(k = 1) 5^(2 k − 1))$] #math.equation(block: true, alt: "4 cubed plus 4 to the power 5 plus 4 to the power 7 plus 4 to the power 9 plus 4 to the power 11")[$4^(3) + 4^(5) + 4^(7) + 4^(9) + 4^(11)$] #math.equation(block: true, alt: "1 plus 4 plus 9 plus 16 plus 25")[$1 + 4 + 9 + 16 + 25$] #math.equation(block: true, alt: "∑ k equals 15 k squared")[$display(limits(∑)^(5)_(k = 1) k^(2))$] #math.equation(block: true, alt: "1 plus 8 plus 27 plus 64 plus 125")[$1 + 8 + 27 + 64 + 125$] #math.equation(block: true, alt: "the fraction 1 over 2 plus the fraction 2 over 3 plus the fraction 3 over 4 plus the fraction 4 over 5 plus the fraction 5 over 6")[$display(frac(1, 2)) + display(frac(2, 3)) + display(frac(3, 4)) + display(frac(4, 5)) + display(frac(5, 6))$] #math.equation(block: true, alt: "∑ k equals 15 the fraction k over k plus 1")[$display(limits(∑)^(5)_(k = 1) display(frac(k, k + 1)))$] #math.equation(block: true, alt: "the fraction 2 over 1 plus the fraction 3 over 2 plus the fraction 4 over 3 plus the fraction 5 over 4 plus the fraction 6 over 5")[$display(frac(2, 1)) + display(frac(3, 2)) + display(frac(4, 3)) + display(frac(5, 4)) + display(frac(6, 5))$] #math.equation(block: true, alt: "the fraction 1 over 1 plus the fraction 2 over 3 plus the fraction 3 over 5 plus the fraction 4 over 7 plus the fraction 5 over 9 plus the fraction 6 over 11")[$display(frac(1, 1)) + display(frac(2, 3)) + display(frac(3, 5)) + display(frac(4, 7)) + display(frac(5, 9)) + display(frac(6, 11))$] #math.equation(block: true, alt: "∑ k equals 16 the fraction k over 2 k minus 1")[$display(limits(∑)^(6)_(k = 1) display(frac(k, 2 k − 1)))$] #math.equation(block: true, alt: "the fraction 3 over 1 plus the fraction 5 over 3 plus the fraction 7 over 5 plus the fraction 9 over 7 plus the fraction 11 over 9")[$display(frac(3, 1)) + display(frac(5, 3)) + display(frac(7, 5)) + display(frac(9, 7)) + display(frac(11, 9))$] #math.equation(block: true, alt: "the fraction 1 over 1 plus the fraction 2 over 2 plus the fraction 4 over 3 plus the fraction 8 over 4 plus ⋯")[$display(frac(1, 1)) + display(frac(2, 2)) + display(frac(4, 3)) + display(frac(8, 4)) + ⋯$] #math.equation(block: true, alt: "∑ k equals 1 ∞ the fraction 2 to the power k minus 1 over k")[$display(limits(∑)^(upright(∞))_(k = 1) display(frac(2^(k − 1), k)))$] #math.equation(block: true, alt: "the fraction 1 over 2 plus the fraction 3 over 4 plus the fraction 9 over 6 plus the fraction 27 over 8 plus ⋯")[$display(frac(1, 2)) + display(frac(3, 4)) + display(frac(9, 6)) + display(frac(27, 8)) + ⋯$] For Problems 21–40, identify the series as arithmetic, geometric or neither, then evaluate it. #math.equation(block: true, alt: "∑ i equals 16 open parenthesis i squared plus 1 close parenthesis")[$display(limits(∑)^(6)_(i = 1) ( i^(2) + 1 ))$] Neither; 97 #math.equation(block: true, alt: "∑ i equals 153 i squared")[$display(limits(∑)^(5)_(i = 1) 3 i^(2))$] #math.equation(block: true, alt: "∑ j equals 14 the fraction 1 over j")[$display(limits(∑)^(4)_(j = 1) display(frac(1, j)))$] Neither; #math.equation(block: false, alt: "the fraction 25 over 12")[$display(frac(25, 12))$] #math.equation(block: true, alt: "∑ j equals 04 the fraction 2 over j plus 1")[$display(limits(∑)^(4)_(j = 0) display(frac(2, j + 1)))$] #math.equation(block: true, alt: "∑ k equals 11001")[$display(limits(∑)^(100)_(k = 1) 1)$] Neither; 100 #math.equation(block: true, alt: "∑ k equals 13003")[$display(limits(∑)^(300)_(k = 1) 3)$] #math.equation(block: true, alt: "∑ q equals 1203 to the power q")[$display(limits(∑)^(20)_(q = 1) 3^(q))$] Geometric; #math.equation(block: false, alt: "5 , 230 , 176 , 600")[$5 , 230 , 176 , 600$] #math.equation(block: true, alt: "∑ p equals 1302 to the power p")[$display(limits(∑)^(30)_(p = 1) 2^(p))$] #math.equation(block: true, alt: "∑ k equals 1200 k")[$display(limits(∑)^(200)_(k = 1) k)$] Atithmetic; #math.equation(block: false, alt: "20 , 100")[$20 , 100$] #math.equation(block: true, alt: "∑ k equals 1150 k")[$display(limits(∑)^(150)_(k = 1) k)$] #math.equation(block: true, alt: "∑ n equals 16 n cubed")[$display(limits(∑)^(6)_(n = 1) n^(3))$] Neither; 441 #math.equation(block: true, alt: "∑ n equals 17 n squared")[$display(limits(∑)^(7)_(n = 1) n^(2))$] #math.equation(block: true, alt: "∑ n equals 030 open parenthesis 3 n minus 1 close parenthesis")[$display(limits(∑)^(30)_(n = 0) ( 3 n − 1 ))$] Arithemetic; 1364 #math.equation(block: true, alt: "∑ k equals 020 open parenthesis 5 k plus 2 close parenthesis")[$display(limits(∑)^(20)_(k = 0) ( 5 k + 2 ))$] #math.equation(block: true, alt: "∑ k equals 025 open parenthesis 5 minus 2 k close parenthesis")[$display(limits(∑)^(25)_(k = 0) ( 5 − 2 k ))$] Arithmetic; #math.equation(block: false, alt: "minus 520")[$− 520$] #math.equation(block: true, alt: "∑ p equals 015 open parenthesis 2 minus 3 p close parenthesis")[$display(limits(∑)^(15)_(p = 0) ( 2 − 3 p ))$] #math.equation(block: true, alt: "∑ j equals 0105 times 2 to the power j")[$display(limits(∑)^(10)_(j = 0) 5 ⋅ 2^(j))$] Geometric; #math.equation(block: false, alt: "24 , 414 , 062")[$24 , 414 , 062$] #math.equation(block: true, alt: "∑ j equals 0103 times 2 to the power j")[$display(limits(∑)^(10)_(j = 0) 3 ⋅ 2^(j))$] #math.equation(block: true, alt: "∑ m equals 01250 open parenthesis 1.08 close parenthesis to the power m")[$display(limits(∑)^(12)_(m = 0) 50 ( 1.08 )^(m))$] Geometric; #math.equation(block: false, alt: "104.76")[$104.76$] #math.equation(block: true, alt: "∑ m equals 018300 open parenthesis 1.12 close parenthesis to the power m")[$display(limits(∑)^(18)_(m = 0) 300 ( 1.12 )^(m))$] For Problems 41–48, + Make a table of values showing the first five partial sums of the series. + Evaluate the series algebraically. #math.equation(block: true, alt: "∑ n equals 1 ∞ open parenthesis the fraction 1 over 2 close parenthesis to the power n")[$display(limits(∑)^(upright(∞))_(n = 1) attach(( display(frac(1, 2)) ), t: n))$] #math.equation(block: true, alt: "1")[$1$] #math.equation(block: true, alt: "∑ n equals 1 ∞ open parenthesis the fraction 1 over 3 close parenthesis to the power n")[$display(limits(∑)^(upright(∞))_(n = 1) attach(( display(frac(1, 3)) ), t: n))$] #math.equation(block: true, alt: "∑ k equals 0 ∞ 12 open parenthesis 0.15 close parenthesis to the power k minus 1")[$display(limits(∑)^(upright(∞))_(k = 0) 12 ( 0.15 )^(k − 1))$] #math.equation(block: true, alt: "14.12")[$14.12$] #math.equation(block: true, alt: "∑ k equals 0 ∞ 25 open parenthesis 0.08 close parenthesis to the power k minus 1")[$display(limits(∑)^(upright(∞))_(k = 0) 25 ( 0.08 )^(k − 1))$] #math.equation(block: true, alt: "∑ j equals 0 ∞ 4 open parenthesis the fraction minus 3 over 5 close parenthesis to the power j")[$display(limits(∑)^(upright(∞))_(j = 0) 4 attach(( display(frac(− 3, 5)) ), t: j))$] #math.equation(block: true, alt: "the fraction 5 over 2")[$display(frac(5, 2))$] #math.equation(block: true, alt: "∑ j equals 0 ∞ 6 open parenthesis the fraction minus 2 over 5 close parenthesis to the power j")[$display(limits(∑)^(upright(∞))_(j = 0) 6 attach(( display(frac(− 2, 5)) ), t: j))$] #math.equation(block: true, alt: "∑ n equals 4 ∞ 3 open parenthesis the fraction 1 over 2 close parenthesis to the power n")[$display(limits(∑)^(upright(∞))_(n = 4) 3 attach(( display(frac(1, 2)) ), t: n))$] #math.equation(block: true, alt: "the fraction 3 over 8")[$display(frac(3, 8))$] #math.equation(block: true, alt: "∑ n equals 3 ∞ 2 open parenthesis the fraction 1 over 3 close parenthesis to the power n")[$display(limits(∑)^(upright(∞))_(n = 3) 2 attach(( display(frac(1, 3)) ), t: n))$] For Problems 49–56, find a fraction equivalent to the repeating decimal. #math.equation(block: true, alt: "0.4 bar")[$0. limits(4)^(―)$] #math.equation(block: true, alt: "the fraction 4 over 9")[$display(frac(4, 9))$] #math.equation(block: true, alt: "0.6 bar")[$0. limits(6)^(―)$] #math.equation(block: true, alt: "0.31 bar")[$0. limits(31)^(―)$] #math.equation(block: true, alt: "the fraction 31 over 99")[$display(frac(31, 99))$] #math.equation(block: true, alt: "0.45 bar")[$0. limits(45)^(―)$] #math.equation(block: true, alt: "0.410 bar")[$0. limits(410)^(―)$] #math.equation(block: true, alt: "2 the fraction 410 over 999")[$2 display(frac(410, 999))$] #math.equation(block: true, alt: "0.027 bar")[$0. limits(027)^(―)$] #math.equation(block: true, alt: "0.128 bar")[$0.12 limits(8)^(―)$] #math.equation(block: true, alt: "the fraction 29 over 225")[$display(frac(29, 225))$] #math.equation(block: true, alt: "0.83 bar")[$0.8 limits(3)^(―)$] The arc length through which the bob of a pendulum moves is nine-tenths of its preceding arc length. Approximately how far will the bob move before coming to rest if the first arc length is #math.equation(block: false, alt: "12")[$12$] inches? 120 in A force is applied to a particle moving in a straight line in such a fashion that each second it moves only one-half of the distance it moved the preceding second. If the particle moves #math.equation(block: false, alt: "10")[$10$] centimeters the first second, approximately how far will it move before coming to rest? A ball returns two-thirds of its preceding height on each bounce. If the ball is dropped from a height of #math.equation(block: false, alt: "6")[$6$] feet, approximately what is the total distance the ball travels before coming to rest? (Hint: Compute separately the total distance the ball falls from the total distance it moves upwards.) 30 ft If a ball is dropped from a height of #math.equation(block: false, alt: "10")[$10$] feet and returns three-fifths of its preceding height on each bounce, approximately what is the total distance the ball travels before coming to rest? (See the Hint for Problem 59.)