Mathematics
Infinite geometric series
An infinite geometric series is a sum of an infinite number of terms that form a geometric sequence. In this series, each term is obtained by multiplying the previous term by a constant ratio. The sum of an infinite geometric series can be calculated using the formula S = a / (1 - r), where S is the sum, a is the first term, and r is the common ratio.
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10 Key excerpts on "Infinite geometric series"
- Bernhard W. Bach, Jr.(Authors)
- 2018(Publication Date)
- Cambridge University Press(Publisher)
A geometric series is constructed from the terms of a geometric sequence and is written as a þ ar þ ar 2 þ ar 3 þ … þ ar n þ … ¼ X ∞ n ¼ 1 a n r n 1 : ð 2 : 2 Þ Many in fi nite series do not have a sum. That is, the in fi nite series or in fi nite summation does not converge to a fi nite numerical value. If an in fi nite series has a de fi nite value, it is called convergent ; otherwise, it is called divergent . In applied problems, we are generally seeking a numer-ical value, for example: How long will it take for the value of an invest-ment to double? How many times do I need to wash a container in order to remove 99% of the unwanted solution? We are interested in solving problems using methods that will converge to a solution or an approxima-tion to the solution. For this reason, this text will focus almost exclusively on convergent series and leave divergent series to mathematicians. One exception will be semiconvergent or asymptotic series , which have appli-cations in physics and computing. 2.2 Convergence and the Sequence of Partial Sums While the concept of an in fi nite summation appears straightforward, there is a subtlety in that an in fi nite summation cannot be evaluated by simply adding consecutive terms. To better understand this point, consider a fi nite summation. A fi nite summation is calculated by simply adding consecutive terms. If there are too many terms to be summed by hand, we can use a computer to complete the 26 In fi nite Series summation. The dif fi culty with an in fi nite summation is that no matter how many thousands of terms are summed, in fi nitely many terms remain to be added. So a direct algebraic approach cannot work, even if it is executed by a computer. Some method other than consecutive addition must therefore be used to determine the sum of an in fi nite series (if it has one). One such method is the method of partial sums . A sequence of partial sums { S n } is constructed from the terms of the in fi nite summation.
eBook - PDFA Radical Approach to Real Analysis
Second Edition
- David Bressoud(Author)
- 2006(Publication Date)
- American Mathematical Society(Publisher)
Any series such as this for which there is a constant ratio between successive summands is called a geometric series . For many values of x , the Infinite geometric series can be summed using the identity 1 + x + x 2 + x 3 + x 4 + · · · = 1 1 − x . (2.7) Examples of this are 1 + 1 3 + 1 9 + 1 27 + 1 81 + · · · = 3 2 = 1 1 − 1 / 3 and 1 − 1 2 + 1 4 − 1 8 + 1 16 − · · · = 2 3 = 1 1 − ( − 1 / 2) . One has to be very careful with equation (2.7). If we set x = 2, we get a very strange equality: 1 + 2 + 4 + 8 + 16 + · · · = 1 1 − 2 = − 1 . (2.8) We need to decide what we mean by an infinite summation. We could define 1 + x + x 2 + x 3 + · · · to mean 1 / (1 − x ), in which case equation (2.8) is correct. We would be in good company. Leonhard Euler accepted this definition. It yields many other interesting 18 2 Infinite Summations results, for example: 1 − 2 + 4 − 8 + 16 − · · · = 1 1 − ( − 2) = 1 3 . In the exhaustive and fascinating account, Convolutions in French Mathematics, 1800– 1840 , Ivor Grattan-Guinness writes, “Some modern appraisals of the cavalier style of 18th-century mathematicians in handling infinite series convey the impression that these poor men set their brains aside when confronted by them.” They did not. Certainly Euler had not set his brain aside. He rather viewed infinite series in a larger context, a context that he makes clear in his article “On divergent series” published in 1760. Euler illustrates his understanding with the series 1 − 1 + 1 − 1 + · · · which he asserts to be equal to 1/2, obtained by setting x = − 1 in equation (2.7). Notable enough, however, are the controversies over the series 1 − 1 + 1 − 1 + 1 − etc . whose sum was given by Leibniz as 1/2, although others dis-agree .
eBook - PDF- Russell L. Herman(Author)
- 2016(Publication Date)
- CRC Press(Publisher)
In this section we look at the special case of a geometric series. A geometric series is of the form ∞ ∑ n = 0 ar n = a + ar + ar 2 + . . . + ar n + . . . . ( 1 . 5 ) 6 an introduction to fourier analysis Here a is the first term and r is called the “ratio.” It is called the ratio because the ratio of two consecutive terms in the sum is r . Example 1 . 6 . For example, 1 + 1 2 + 1 4 + 1 8 + . . . is an example of a geometric series. We can write this using summa-tion notation, 1 + 1 2 + 1 4 + 1 8 + . . . = ∞ ∑ n = 0 1 1 2 n . Thus, a = 1 is the first term and r = 1 2 is the common ratio of succes-sive terms. Next, we seek the sum of this infinite series, if it exists. The sum of a geometric series, when it exists, can easily be determined. We consider the n th partial sum: s n = a + ar + . . . + ar n -2 + ar n -1 . ( 1 . 6 ) Now, multiply this equation by r . rs n = ar + ar 2 + . . . + ar n -1 + ar n . ( 1 . 7 ) Subtracting these two equations, while noting the many cancelations, we have ( 1 -r ) s n = ( a + ar + . . . + ar n -2 + ar n -1 ) -( ar + ar 2 + . . . + ar n -1 + ar n ) = a -ar n = a ( 1 -r n ) . ( 1 . 8 ) Thus, the n th partial sums can be written in the compact form 4 4 We have obtained the sum of a geomet-ric progression, N -1 ∑ n = 0 ar n = a ( 1 -r n ) 1 -r . s n = a ( 1 -r n ) 1 -r . ( 1 . 9 ) The sum of the geometric series, if it exists, is given by S = lim n → ∞ s n . Letting n get large in the partial sum ( 1 . 9 ), we need only evaluate lim n → ∞ r n . From the special limits in the Appendix we know that this limit is zero for | r | < 1. Thus, we have Geometric Series The sum of the geometric series exists for | r | < 1 and is given by ∞ ∑ n = 0 ar n = a 1 -r , | r | < 1. ( 1 . 10 ) The reader should verify that the geometric series diverges for all other values of r . Namely, consider what happens for the separate cases | r | > 1, r = 1 and r = -1. Next, we present a few typical examples of geometric series.
- Mary L. Boas(Author)
- 2011(Publication Date)
- Wiley(Publisher)
C H A P T E R 1 Infinite Series, Power Series 1. THE GEOMETRIC SERIES As a simple example of many of the ideas involved in series, we are going to consider the geometric series. You may recall that in a geometric progression we multiply each term by some fixed number to get the next term. For example, the sequences 2 , 4 , 8 , 16 , 32 , . . . , (1.1a) 1 , 2 3 , 4 9 , 8 27 , 16 81 , . . . , (1.1b) a, ar, ar 2 , ar 3 , . . . , (1.1c) are geometric progressions. It is easy to think of examples of such progressions. Suppose the number of bacteria in a culture doubles every hour. Then the terms of (1.1a) represent the number by which the bacteria population has been multiplied after 1 hr, 2 hr, and so on. Or suppose a bouncing ball rises each time to 2 3 of the height of the previous bounce. Then (1.1b) would represent the heights of the successive bounces in yards if the ball is originally dropped from a height of 1 yd. In our first example it is clear that the bacteria population would increase with-out limit as time went on (mathematically, anyway; that is, assuming that nothing like lack of food prevented the assumed doubling each hour). In the second example, however, the height of bounce of the ball decreases with successive bounces, and we might ask for the total distance the ball goes. The ball falls a distance 1 yd, rises a distance 2 3 yd and falls a distance 2 3 yd, rises a distance 4 9 yd and falls a distance 4 9 yd, and so on. Thus it seems reasonable to write the following expression for the total distance the ball goes: (1.2) 1 + 2 · 2 3 + 2 · 4 9 + 2 · 8 27 + · · · = 1 + 2 ( 2 3 + 4 9 + 8 27 + · · · ) , where the three dots mean that the terms continue as they have started (each one being 2 3 the preceding one), and there is never a last term. Let us consider the expression in parentheses in (1.2), namely (1.3) 2 3 + 4 9 + 8 27 + · · · . 1
eBook - PDF- Hugo D. Junghenn(Author)
- 2015(Publication Date)
- Chapman and Hall/CRC(Publisher)
Chapter 6 Numerical Infinite Series An infinite series is the limit of a sequence of expanding finite sums. The terms of these sums may be real numbers or functions. In this chapter we examine the convergence behavior of series of the former type; series whose terms are functions are treated in the next chapter. In the first section, we give examples of series that may be summed, that is, for which an explicit numerical value may be calculated. The remaining sections describe various tests for convergence of general series. Additional methods of summing series may be found in Section 7.4. 6.1 Definition and Examples 6.1.1 Definition. Let { a n } be a sequence of real numbers. The various symbols a n = n a n = ∞ n =1 a n = a 1 + a 2 + · · · + a n + · · · represent what is called an infinite series with n th term a n or, simply, a series . The n th partial sum of the series is defined by s n = n k =1 a k . The series is said to converge if the sequence of partial sums converges, in which case we write a n = lim n s n and call ∑ a n the sum of the series . If the sequence { s n } diverges, then the series is said to diverge . ♦ 6.1.2 Remark. A series may begin with an index other than 1. In this regard, note that, because s n = s m -1 + n k = m a k , n ≥ m > 1 , 163 164 A Course in Real Analysis the series s := ∑ ∞ n =1 a n converges iff ∑ ∞ n = m a n converges. In this case the “tail end” of the series tends to zero: lim m → + ∞ ∞ n = m a n = lim m → + ∞ ( s -s m -1 ) = 0 . ♦ 6.1.3 Example. Using the definition e := lim n →∞ (1 + 1 /n ) n (see 2.2.4), we show that e = ∞ n =0 1 n ! . First, since the partial sums s n := ∑ n k =0 1 /k ! increase, the limit s := lim n s n exists in R . From the calculations in 2.2.4, (1 + 1 /n ) n = 2 + n k =2 1 k ! (1 -1 /n )(1 -2 /n ) · · · (1 -( k -1) /n ) ≤ s n . Letting n → ∞ , we obtain e ≤ s . On the other hand, if n > m , then (1 + 1 /n ) n > 2 + m k =2 1 k ! (1 -1 /n )(1 -2 /n ) · · · (1 -( k -1) /n ) .
eBook - PDF- Sheldon Axler(Author)
- 2011(Publication Date)
- Wiley(Publisher)
Following the usual procedure for infinite sums, first we evaluate the partial sums n k=1 (-1) k , getting n k=1 (-1) k = -1 if n is odd 0 if n is even. Thus the sequence of partial sums is the alternating sequence of -1’s and 0’s. This sequence of partial sums does not have a limit. Thus the infinite sum is undefined. section 8.3 Limits 489 We turn now to the problem of finding a formula for evaluating an Infinite geometric series. Fix a number r = 1, and consider the geometric series 1 + r + r 2 + r 3 + · · · ; here the ratio of consecutive terms is r . The sum of the first n terms is 1 + r + r 2 + · · · + r n-1 . The term following the last term would be r n ; thus by our formula for evaluating a geometric series we have 1 + r + r 2 + · · · + r n-1 = 1 - r n 1 - r . By definition, the infinite sum 1 + r + r 2 + r 3 + · · · equals the limit (if it exists) of the partial sums above as n goes to infinity. We already know that the limit of r n as n goes to infinity is 0 if |r | < 1 (and does not exist if |r | > 1). Thus we get the following beautiful formula: If |r | ≥ 1, then this infinite sum is not defined. Evaluating an Infinite geometric series If |r | < 1, then 1 + r + r 2 + r 3 + · · · = 1 1 - r . Any Infinite geometric series can be reduced to the form above by factoring out the first term. The following example illustrates the procedure. example 9 Evaluate the geometric series 7 3 + 7 9 + 7 27 + · · · . solution We factor out the first term 7 3 and then apply the formula above, getting 7 3 + 7 9 + 7 27 + · · · = 7 3 1 + 1 3 + 1 9 + · · · = 7 3 · 1 1 - 1 3 = 7 2 . Decimals as Infinite Series A digit is one of the numbers 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. Each real number t between 0 and 1 can be expressed as a decimal in the form t = 0.d 1 d 2 d 3 . . . , where d 1 , d 2 , d 3 , . . . is a sequence of digits. The interpretation of this repre- sentation is that t = d 1 10 + d 2 100 + d 3 1000 + · · · , which we can write in summation notation as
eBook - PDF- Ajit Kumar, S. Kumaresan(Authors)
- 2014(Publication Date)
- Chapman and Hall/CRC(Publisher)
33333 . . . . 153 154 CHAPTER 5. INFINITE SERIES 5.1 Convergence and Sum of an Infinite Series Definition 5.1.1. Given a sequence ( a n ) of real numbers, a formal sum of the form ∑ ∞ n =1 a n (or ∑ a n , for short) is called an infinite series. For any n ∈ N , the finite sum s n := a 1 + · · · + a n is called the ( n -th) partial sum of the series ∑ a n . A more formal definition of an infinite series is as follows. By the symbol ∑ n a n we mean the sequence ( s n ) where s n := a 1 + · · · + a n . We say that the infinite series ∑ a n is convergent if the sequence ( s n ) of partial sums is convergent. In such a case, the limit s := lim s n is called the sum of the series and we denote this fact by the symbol ∑ a n = s . We say that the series ∑ a n is divergent if the sequence of its partial sums is divergent. The series ∑ n a n is said to be absolutely convergent if the infinite series ∑ n | a n | is convergent. Note that a series ∑ a n of non-negative terms, (that is, a n ≥ 0 for all n ) is convergent iff it is absolutely convergent. If a series is convergent but not absolutely convergent, then it is said to be conditionally convergent . Let us look at some examples of series and their convergence. Example 5.1.2. Let ( a n ) be a constant sequence a n = c for all n . Then the infinite series ∑ a n is convergent iff c = 0. For, the partial sum is s n = nc . Thus ( s n ) is convergent iff c = 0. (Why? Use the Archimedean property.) Example 5.1.3. Let a n be non-negative real numbers and assume that ∑ a n is convergent. Since s n +1 = s n + a n +1 , it follows that the sequence ( s n ) is increasing. We have seen (Theorem 2.3.2) that ( s n ) is convergent iff it is bounded above. Hence a series of non-negative terms is convergent iff the sequence of partial sums is bounded. Note that if ∑ a n is convergent, then ∑ a n = lub { s n : n ∈ N } . Example 5.1.4 (Geometric Series) . This is the most important example.
eBook - PDFAlgebra
A Combined Course 2E
- Charles P. McKeague(Author)
- 2018(Publication Date)
- XYZ Textbooks(Publisher)
A calculator will give the result as 147,620. D Infinite geometric series Suppose the common ratio for a geometric sequence is a number whose absolute value is less than 1—for instance, 1 _ 2 . The sum of the first n terms is given by the formula S n = a 1 1 __ 2 n − 1 __ 1 __ 2 − 1 As n becomes larger and larger, the term 1 _ 2 n will become closer and closer to 0. That is, for n = 10, 20, and 30, we have the following approximations. 1 __ 2 10 ≈ 0.001 1 __ 2 20 ≈ 0.000001 1 __ 2 30 ≈ 0.000000001 6. Find the sum of the first 9 terms of 8, 24, 72, 216,… Answer 6. 78,728 Chapter 15 Sequences and Series 1154 Therefore, for large values of n, there is little difference between the expression a 1 (r n − 1) ________ r − 1 and the expression a 1 (0 − 1) _______ r − 1 = −a 1 ____ r − 1 = a 1 ____ 1 − r if | r | < 1 In fact, the sum of the terms of a geometric sequence in which | r | < 1 actually becomes the expression a 1 ____ 1 − r as n approaches infinity. To summarize, we have the following: PROPERTY The Sum of an Infinite Geometric Sequence If a geometric sequence has first term a 1 and common ratio r such that | r | < 1, then the following is called an Infinite geometric series: S = ∑ i=0 ∞ a 1 r i = a 1 + a 1 r + a 1 r 2 + a 1 r 3 + ⋅ ⋅ ⋅ Its sum is given by the formula S = a 1 _ 1 − r EXAMPLE 7 Find the sum of the Infinite geometric series 1 __ 5 + 1 __ 10 + 1 __ 20 + 1 __ 40 + ⋅ ⋅ ⋅ Solution The first term is a 1 = 1 _ 5 , and the common ratio is r = 1 _ 2 , which has an absolute value less than 1. Therefore, the sum of this series is S = a 1 _ 1 − r = 1 _ 5 _ 1 − 1 _ 2 = 1 _ 5 _ 1 _ 2 = 2 __ 5 EXAMPLE 8 Show that 0.999 . . . is equal to 1. Solution We begin by writing 0.999 .
eBook - PDF- Sheldon Axler(Author)
- 2011(Publication Date)
- Wiley(Publisher)
Following the usual procedure for infinite sums, first we evaluate the partial sums n X k=1 (-1) k , getting n X k=1 (-1) k = -1 if n is odd 0 if n is even. Thus the sequence of partial sums is the alternating sequence of -1’s and 0’s. This sequence of partial sums does not have a limit. Thus the infinite sum is undefined. section 8.3 Limits 489 We turn now to the problem of finding a formula for evaluating an Infinite geometric series. Fix a number r 6= 1, and consider the geometric series 1 + r + r 2 + r 3 + · · · ; here the ratio of consecutive terms is r . The sum of the first n terms is 1 + r + r 2 + · · · + r n-1 . The term following the last term would be r n ; thus by our formula for evaluating a geometric series we have 1 + r + r 2 + · · · + r n-1 = 1 - r n 1 - r . By definition, the infinite sum 1 + r + r 2 + r 3 + · · · equals the limit (if it exists) of the partial sums above as n goes to infinity. We already know that the limit of r n as n goes to infinity is 0 if |r | < 1 (and does not exist if |r | > 1). Thus we get the following beautiful formula: If |r | ≥ 1, then this infinite sum is not defined. Evaluating an Infinite geometric series If |r | < 1, then 1 + r + r 2 + r 3 + · · · = 1 1 - r . Any Infinite geometric series can be reduced to the form above by factoring out the first term. The following example illustrates the procedure. example 9 Evaluate the geometric series 7 3 + 7 9 + 7 27 + · · · . solution We factor out the first term 7 3 and then apply the formula above, getting 7 3 + 7 9 + 7 27 + · · · = 7 3 1 + 1 3 + 1 9 + · · · = 7 3 · 1 1 - 1 3 = 7 2 . Decimals as Infinite Series A digit is one of the numbers 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. Each real number t between 0 and 1 can be expressed as a decimal in the form t = 0.d 1 d 2 d 3 . . . , where d 1 , d 2 , d 3 , . . . is a sequence of digits. The interpretation of this repre- sentation is that t = d 1 10 + d 2 100 + d 3 1000 + · · · , which we can write in summation notation as
eBook - PDFCalculus
Concepts and Contexts, Enhanced Edition
- James Stewart(Author)
- 2018(Publication Date)
- Cengage Learning EMEA(Publisher)
So, by taking the sum of sufficiently many terms, we can get as close as we like to the number . The table shows the first ten partial sums and the graph in Figure 2 shows how the sequence of partial sums approaches . 3 s n 3 3 Copyright 2019 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 568 CHAPTER 8 INFINITE SEQUENCES AND SERIES Is the series convergent or divergent? SOLUTION Let’s rewrite the n th term of the series in the form : We recognize this series as a geometric series with and . Since , the series diverges by (4). Expressing a repeating decimal as a rational number Write the number . . . as a ratio of integers. SOLUTION After the first term we have a geometric series with and . Therefore A series with variable terms Find the sum of the series , where SOLUTION Notice that this series starts with and so the first term is . (With series, we adopt the convention that even when .) Thus This is a geometric series with and . Since , it converges and (4) gives A telescoping sum Show that the series is convergent, and find its sum. SOLUTION This is not a geometric series, so we go back to the definition of a convergent series and compute the partial sums . s n n i 1 1 i i 1 1 1 2 1 2 3 1 3 4 1 n n 1 n 1 1 n n 1 EXAMPLE 6 n 0 x n 1 1 x 5 r x 1 r x a 1 n 0 x n 1 x x 2 x 3 x 4 x 0 x 0 1 x 0 1 n 0 x 1. n 0 x n EXAMPLE 5 23 10 17 990 1147 495 2.317 2.3 17 10 3 1 1 10 2 2.3 17 1000 99 100 r 1 10 2 a 17 10 3 2.3171717.
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