# 16.1. Sequences and Series. Introduction. Prerequisites. Learning Outcomes. Learning Style

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1 Sequences and Series 16.1 Introduction In this block we develop the ground work for later blocks on infinite series and on power series. We begin with simple sequences of numbers and with finite series of numbers. We introduce the summation notation for the description of series. Finally, we consider arithmetic and geometric series and obtain expressions for the sum of n terms of both types of series. Prerequisites Before starting this Block you should... Learning Outcomes After completing this Block you should be able to... check if a sequence of numbers is convergent use the summation notation to specify series recognise arithmetic and geometric series and find their sums 1 understand and be able to use the basic rules of algebra 2 be able to find limits of algebraic expressions Learning Style To achieve what is expected of you... allocate sufficient study time briefly revise the prerequisite material attempt every guided exercise and most of the other exercises

2 1. Introduction A sequence is any succession of numbers. For example the sequence 1, 1, 2, 3, 5, 8,... which is known as the Fibonacci sequence, is formed by adding two consecutive terms together to obtain the next term. The numbers in this sequence continually increase without bound and we say this sequence diverges. An example of a convergent sequence is the harmonic sequence 1 1, 2, 1 3, 1 4,... Here we see the magnitude of these numbers continually decrease and it is obvious that the sequence converges to the number zero. The related alternating harmonic sequence 1, 1 2, 1 3, 1 4,... is also convergent to the number zero. Whether or not a sequence is convergent is often easy to deduce by graphing the individual terms. The following diagrams show how the individual terms of the harmonic and alternating harmonic series behave as the number of terms increase. 1 1/2 1/3 1/4 harmonic term in sequence 1 alternating harmonic 1/3 1/4 1/ term in sequence Now do this exercise Graph the sequence: Is this convergent? A general sequence is denoted by 1, 1, 1, 1,... a 1,a 2,..., a n,... Answer in which a 1 is the first term, a 2 is the second term and a n is the n th term is the sequence. For example, in the harmonic sequence a 1 =1, a 2 = 1 2,...,a n = 1 n Engineering Mathematics: Open Learning Unit Level 1 2

3 whilst for the alternating harmonic sequence the n th term is: a n = ( 1)n+1 n in which ( 1) n =+1ifn is an even number and ( 1) n = 1 ifn is an odd number. Definition The sequence a 1,a 2,...,a n,... is said to be convergent if the limit of a n as n increases can be found. (Mathematically we say that lim a n exists). n If the sequence is not convergent it is said to be divergent Try each part of this exercise Verify that the sequence 3 1 2, 4 2 3, 5 3 4,... is convergent. Part (a) First find the expression for the n th term. Answer Part (b) Now find the limit of a n as n increases. Answer 2. Arithmetic and Geometric Progressions Consider the sequences: 1, 4, 7, 10,... and 3, 1, 1, 3,... In both, any particular term is obtained from the previous term by the addition of a constant value (3 and 2 respectively). Each of these sequences are said to be arithmetic sequences or arithmetic progressions and have general form: a, a + d, a +2d, a +3d,..., a+(n 1)d,... in which a, d are given numbers. In the first example above a =1,d= 3 whereas, in the second example, a =3,d= 2. The difference between any two terms of a given arithmetic sequence gives the value of d. Two sequences which are not arithmetic sequences are: 1, 2, 4, 8,... 3 Engineering Mathematics: Open Learning Unit Level 1

4 1, 1 3, 1 9, 1 27,... In each case a particular term is obtained from the previous term by multiplying by a constant factor (2 and 1 respectively). Each are examples of geometric sequences or geometric 3 progressions with the general form: a, ar, ar 2,ar 3,... where a is the first term and r is called the common ratio. In the two examples above a =1, r = 2 in the first sequence and a = 1, r = 1 in the second. 3 Try each part of this exercise Part (a) Find a, d for the arithmetic sequence 3, 9, 15,... Answer Part (b) Find a, r for the geometric sequence 8, 8 7, 8 49,... Answer Part (c) Write out the first four terms of the geometric series with a =4,r = 2.. Answer The reader should note that many sequences (for example the harmonic sequences) are neither arithmetic or geometric. 3. Series A series is the sum of the terms of a sequence. For example, the harmonic series is whilst the alternating harmonic series is The Summation Notation If we consider a general sequence a 1,a 2,...,a n,... then the sum of the first k terms a 1 + a 2 + a a k is concisely denoted by That is, a 1 + a 2 + a a k = k a p k a p. Engineering Mathematics: Open Learning Unit Level 1 4

5 k When we encounter the expression a p we let the index p in the term a p take, in turn, the values 1, 2,...,k and then add all these terms together. So, for example 3 a p = a 1 + a 2 + a 3 7 a p = a 2 + a 3 + a 4 + a 5 + a 6 + a 7 p=2 Note that p is a dummy index; any letter could be used as the index. For example 6 a m each represent the same collection of terms, being a 1 + a 2 + a 3 + a 4 + a 5 + a 6. m=1 6 a i, and In order to be able to use this summation notation we need to obtain a suitable expression for the typical term in the series. For example, the finite series k 2 i=1 may be written as In the same way k p 2 since the typical term is clearly p 2 in which p =1, 2, 3,...,k in turn = ( 1) p+1 p since an expression for the typical term in this alternating harmonic series is a p = ( 1)p+1 p. Try each part of this exercise Part (a) Write in summation form the series First find an expression for the typical term, the p th term. Answer Part (b) Write out all the terms of the series 5 ( 1) p (p +1) 2. Answer 5 Engineering Mathematics: Open Learning Unit Level 1

6 4. Summing Series The arithmetic series Consider the finite arithmetic series with 14 terms A simple way of working out the value of the sum is to create a second series which is the first written in reverse order. Thus we have two series, each with the same value A: A = or A = Now, adding the terms of these series in pairs 2A = = 28(14) = 392 so A = 196 We can use this approach to find the sum of n terms of a general arithmetic series. If A =[a]+[a + d]+[a +2d]+...+[a +(n 2)d]+[a +(n 1)d] then again simply writing the terms in reverse order: A =[a +(n 1)d]+[a +(n 2)d]+...+[a +2d]+[a + d]+[a] Adding these two identical equations together we have 2A =[2a +(n 1)d]+[2a +(n 1)d]+...+[2a +(n 1)d] That is, every one of the n terms on the right-hand side has the same value: [2a +(n 1)d]. Hence 2A = n[2a +(n 1)d] so A = 1 n[2a +(n 1)d]. 2 The arithmetic series Key Point having n terms has sum A where: [a]+[a + d]+[a +2d]+...+[a +(n 1)d] A = 1 n[2a +(n 1)d] 2 As an example So has a =1,d=2,n=14 A = = 14 [2 + (13)2] = Engineering Mathematics: Open Learning Unit Level 1 6

7 The geometric series We can also sum a general geometric series. Let G = a + ar + ar ar n 1 be a geometric series having exactly n terms. To obtain the value of G in a more convenient form we first multiply through by the common ratio r: Now, writing the two series together: rg = ar + ar 2 + ar ar n G = a + ar + ar ar n 1 rg = ar + ar 2 + ar ar n 1 + ar n Subtracting the second expression from the first we see that all terms on the right-hand side cancel out, except for the first term of the first expression and the last term of the second expression so that G rg =(1 r)g = a ar n Hence (assuming r 1) G = a(1 rn ) 1 r (Of course, if r = 1 the geometric series simplifies to a simple arithmetic series with d = 0 and has sum G = na) The geometric series having n terms has sum G where Key Point a + ar + ar ar n 1 G = a(1 rn ), if r 1 and G = nr, if r =1 1 r Try each part of this exercise Find the sum of each of the following series: Part (a) Answer Part (b) Answer 7 Engineering Mathematics: Open Learning Unit Level 1

8 More exercises for you to try 1. Which of the following sequences is convergent? (a) sin π 2, sin 2π 2, sin 3π 2, sin 4π 2,... (b) sin π 2 π 2, sin 2π 2 2π 2, sin 3π 2 3π 2, sin 4π 2 4π 2, Write the following series in summation form: (a) ln 1 2 1, ln 3 3 2, ln 5 4 3,..., ln (b) 1 2 (1 + (100) 2 ), 1 3 (1 (200) 2 ), 1 4 (1 + (300) 2 ),..., 1 9 (1 (800) 2 ) 3. Write out the first three terms and the last term of the following series: (a) 17 3 p 1 p!(18 p) (b) 17 p=4 ( p) p+1 p(2 + p) 4. Sum the series: (a) (b) (c) Answer Engineering Mathematics: Open Learning Unit Level 1 8

9 5. Computer Exercise or Activity For this exercise it will be necessary for you to access the computer package DERIVE. Derive is of some help with sequences and series. First it can be used to obtain the limits of expressions in n as n. For example to obtain the limit: lim n n +2 n(n +1) we would key in (n +2)/(n(n + 1)). DERIVE responds with n +2 n (n +1) then hit Calculus:Limit to obtain the Calculus limit screen. In the limit Point box choose (from the list of symbols) and then choose Both in the Approach From section. If you then hit the OK button DERIVE responds with lim n n +2 n (n +1) Finally, hit Simplify:Basic to obtain the response 0 which is as expected. DERIVE will also obtain the sum of finite series. To do this you will need to obtain an expression for the n th term of the series. For example, to find we need to recognise that the n th term is n. Thus = n. To obtain this sum using DERIVE, enter the Calculus:Sum box and enter the expression for the n th term. In this case enter n. Make sure the Variable box contains n. Then choose Definite in the sum box and choose appropriate limits (1 as the Lower limit and 100 as the Upper limit). If you then hit the OK button DERIVE responds 100 n=1 n 9 Engineering Mathematics: Open Learning Unit Level 1 n=1

10 Finally, hit Simplify:Basic to get the value of the sum: 5050 a result which we have already obtained. As an exercise you could experiment with the limit and summation capabilities of DERIVE. DERIVE will also obtain the algebraic expressions for some finite sums: For example if you key in: p 2 then Calculus:Sum as before but this time choosing Variable p and Upper limit n and then hit the OK button. DERIVE responds n p 2 Now hit Simplify:Basic to obtain the expression n (n +1) (2 n +1) 6 which is easily checked to be correct (for e.g = 5(6)(11) =55.) 6 Engineering Mathematics: Open Learning Unit Level 1 10

11 End of Block Engineering Mathematics: Open Learning Unit Level 1

12 term in sequence 1 Not convergent. The terms in the sequence do not converge to a particular value. The value oscillates. Back to the theory Engineering Mathematics: Open Learning Unit Level 1 12

13 a n = n+2 n(n+1) Back to the theory 13 Engineering Mathematics: Open Learning Unit Level 1

14 n +2 n(n +1) = Hence the sequence is convergent. [ 1+ 2 ] n 1 n +1 n +1 0 as n increases Back to the theory Engineering Mathematics: Open Learning Unit Level 1 14

15 a =3, d =6 Back to the theory 15 Engineering Mathematics: Open Learning Unit Level 1

16 a =8, r = 1 7 Back to the theory Engineering Mathematics: Open Learning Unit Level 1 16

17 4, 8, 16, 32,... Back to the theory 17 Engineering Mathematics: Open Learning Unit Level 1

18 a p = 1 p(p+1) = 1 p(p +1) Back to the theory Engineering Mathematics: Open Learning Unit Level 1 18

19 Back to the theory 19 Engineering Mathematics: Open Learning Unit Level 1

20 a =1, d =1, n = = (99) = 50(101) = Back to the theory Engineering Mathematics: Open Learning Unit Level 1 20

21 a = 1 2, r = 1 3, n = = ( 1 ( 1 3) 6) = 3 4 ( 1 ( ) ) = Back to the theory 21 Engineering Mathematics: Open Learning Unit Level 1

22 1(a) no; this sequence is 1, 0, 1, 0, 1,... which does not converge. 1 (b) yes; this sequence is, 0, 1, 0, 1,... which converges to zero. π/2 3π/2 5π/2 14 ln(2p 1) 8 ( 1) p 2. (a) (b) (p + 1)(p) (p + 1)(1 + ( 1) p+1 p ) 3. (a) 1 17, 3 2!(16), 3 2 3!(15),..., ! (b) 45, 5 6, 67,..., (4)(6) (5)(7) (6)(8) (17)(19) 4. (a) This is an arithmetic series with a = 5, d=4,n=9. A =99 (b) This is an arithmetic series with a = 5, d= 4, n=9. A = 189 (c) This is a geometric series with a = 1,r= 1,n=6. G Back to the theory Engineering Mathematics: Open Learning Unit Level 1 22

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