RI Sequences and Series C7A Lect Notes

RAFFLES INSTITUTION H2 Mathematics (9758) 2025 Year 5 ______________________________ Chapter 7A: Sequences and Series Page 1 of 17 Chapter 7A: Sequences and Series SYLLABUS INCLUDES  Concepts of sequence and series for finite and infinite cases  Sequence as function f( )yn where n is a positive integer  Sequence given by a formula for the nth term  Relationship between nu (the nth term) and Sn (the sum to n terms)  Sequence generated by the relation 1 f( )nnuu  , including the use of a graphing calculator to generate the sequence  Sum and difference of two series  Convergence of a series and the sum to infinity. CONTENT 1 Sequences and Series 1.1 Basic Definitions 1.2 Convergence of a Sequence 1.3 Convergence of a Series 1.4 Sequence generated by a Recurrence Relation 2 Representing a Series by ∑ (Sigma) Notation 2.1 Properties of the ∑ (Sigma) Notation 2.2 Summation using Standard Results INTRODUCTION Sequences and series occur frequently in real life. In this chapter, we will learn more about these concepts, and special sequences such as arithmetic and geometric progressions. You will also get to solve practical problems involving loan repayment, compound interest, etc. You may just find yourself using the knowledge and skills when you take up loans to buy a car or a property some years down the road!

Raffles Institution H2 Mathematics 2025 Year 5 _________________________________________________________________________________________ ______________________________ Chapter 7A: Sequences and Series Page 2 of 17 1 Sequences and Series 1.1 Basic Definitions A sequence is a set of numbers arranged in a defined order. Each number in the sequence is called a term of the sequence. A sequence can have finite or infinite number of terms. A sequence can be defined by a formula for the general term  fnun . Some examples: Formula for nu n 6u Sequence Finite / Infinite 1. 2 nun 1, 2,3,...,12n 36 1, 4, 9, 16, 25, … , 144 Finite 2. 20 5( 1)nun  2, 3, 4,...n 5 15,10,5, 0, 5,... Infinite 3. 2 1 2 n nu  1, 2,3,...n 1 16 1112 , 1 , , , , . . .248 Infinite 4. (1 ) n nun 0,1, 2,3,...,50n 6 0, 1, 2, 3, 4, 5,... ,50 Finite 5. 15 15 22 5 nn nu       n  8 Fibonacci Sequence 1, 1, 2, 3, 5, … Infinite

Raffles Institution H2 Mathematics 2025 Year 5 _________________________________________________________________________________________ ______________________________ Chapter 7A: Sequences and Series Page 3 of 17 Example 1 [9740/2009/01/Q1] (i) The first three terms of a sequence are given by 12 310, 6, 5.uu u  Given that nu is a quadratic polynomial in ,n find nu in terms of .n [4] (ii) Find the set of values on n for which nu is greater than 100. [2] Solutions (i) Let un = an2 + bn + c. u1 = a + b + c = 10 u2 = 4a + 2b + c = 6 u3 = 9a + 3b + c = 5 Using GC, a = 1.5, b = –8.5, c = 17.  un = 1.5n2 – 8.5n + 17. (ii) 2 2 100 1.5 8.5 17 100 1.5 8.5 17 100 0 nu nn nn       Let y = 1.5n2 – 8.5n + 17 – 100 From GC, since n  , n > 10.79. Hence solution set = {n  ℤ+ : n  11}. OR 2 2 100 1.5 8.5 17 100 1.5 8.5 17 100 0 nu nn nn       Let y = 1.5n2 – 8.5n + 17 – 100 From G.C., n y 10 18 < 0 11 5 > 0 Hence solution set = {n  ℤ+ : n  11}. GC Keystrokes 1. Press o and key in the expression. 2. Press ys for the table of values. As the question needed the 1Y value to be positive, scroll down until you reach the first x-value for which 1Y is positive.

Raffles Institution H2 Mathematics 2025 Year 5 _________________________________________________________________________________________ ______________________________ Chapter 7A: Sequences and Series Page 4 of 17 A series is the sum of the terms of a sequence and is denoted by 12 1 ...nn nSu u u u   . A series can have finite or infinite number of terms. Note Example 2 [9758/2020/02/Q2c] The sum of the first n terms of a series is 32 11 4 ,nn n where n is a positive integer. (i) Find an expression for the nth term of this series, giving your answer in its simplest form. (ii) The sum of the first m terms of this series, where m > 3, is equal to the sum of the first three terms of this series. Find the value of m. Solutions (i) Let nS denote the sum of the first n terms.          1 3232 3232 33 2 2 2 For 2, term 11 4 1 11 1 4 1 1 11 1 4 1 3 3 1 11 2 1 4 th nnnn S S nn nn n n nn n n n n nnn n n n n                        2 2 = 3 3 1 22 11 4 3 25 16 nn n nn     23 11st term = S 1 11 1 4 6   and it follows the form 232 5 1 6nn when 1.n Thus 2th term = 3 25 16 for 1.nn n n   (ii)   3 232 3 32 11 4 3 11 3 4 3 60 11 4 60 0 2,3 or 10 10 since 3 mSS mm m mm m m mm             In the next section, we will look at convergence of a sequence and a series.  1 for 2nn nuSS n    11uS

Raffles Institution H2 Mathematics 2025 Year 5 _________________________________________________________________________________________ ______________________________ Chapter 7A: Sequences and Series Page 5 of 17 1.2 Convergence of a Sequence A sequence nu is said to be convergent if it approaches a unique value as n approaches infinity. For example, from the previous section, given 2 1 2 n nu  , where 1, 2,3,...n , it creates the sequence as follows: 1112 , 1 , , , , . . .248 . As n , we observed that 2 1 02 n  , hence we can conclude that 0nu  and this sequence converges. 1.3 Convergence of a Series If 12 1 ...nn nSu u u u   approaches a unique value as n approaches infinity, we say that the series converges and that the sum to infinity S exists. Otherwise, the series is said to diverge or that the sum to infinity S does not exist. For example, consider the series 2 121 2 n nS    . It can be proven that 141 2 n nS   (try proving this on your own after you have studied geometric series). We see that as n , 1 02 n  and 4.nS  So, we say that the series 11 12 1 + . . .24 8   converges to 4 and that 4S . Note that from Section 1.2, the corresponding sequence 1112 , 1 , , , , . . .248 converges to 0. Here are some examples of a divergent series. (a) If 1 nu n , the sequence 1111, , , ,234  converges to 0 since as n , 1 0n  . However, the series 1111 234 diverges since 1 11 1111 1 11 111111 2 34 5678 2 44 8888 1111 222                              and 1111 a s 222 n    (b) If  1 n nu  , then 1111 1 n nS        is divergent.

Raffles Institution H2 Mathematics 2025 Year 5 _________________________________________________________________________________________ ______________________________ Chapter 7A: Sequences and Series Page 6 of 17 1 2 3 4 1, 1 2, 1 1 0 3, 1 1 1 1 4 , 1111 0 nS nS nS nS                   As a result, we can conclude that 1 if is oddnSn and 0 if is even.nSn Hence, the series nS does not approach a unique value and is divergent. Note that the sequence 1,1, 1,1,  is also divergent since  1 n nu  does not approach a unique value as n . (c) If lnnun , then both the sequence ln1, ln 2, ln 3, and the corresponding series ln1 ln 2 ln 3nS   is divergent since ln1 ln 2 ln 3 ln ln(1 2 3 ... ) ln !nSn n n      and ln n and ln !n  as n . We can verify the result by using a GC, as shown below. From GC, we see that as n increases, the value o