Series. Definition. a 1 + a 2 + a 3 + is called an infinite series or just series. Denoted by. n=1

Transcription

1 Definition a 1 + a 2 + a 3 + is called an infinite series or just series. Denoted by a n, or a n. Chapter 11: Sequences and, Section / 40

2 Given a series a n. The partial sum is the sum of the first n terms of the series, denoted by s n. s n := n a i = a 1 + a a n. i=1 If lim n s n exists as a finite number, then the series and we say it is convergent. a n := lim n s n, If lim n s n does not exist, we say a n is divergent. Chapter 11: Sequences and, Section / 40

3 3n 2n+3 Example 1. Suppose s n =. Then by definition, a 3n n = lim n s n = lim n 2n+3 = 3 2. Chapter 11: Sequences and, Section / 40

4 Example 2. Show that to 1. Solution: The partial sum Note that 1 n(n+1) s n = n(n + 1). 1 n(n + 1) = 1 n 1 n + 1. So is convergent. And it is equal s n = (1 1 2 ) + ( ) + ( ) + ( 1 n 1 n + 1 ) All the terms, except 1 and 1 n+1, cancel. So s n = 1 1 n+1. Hence lim s n = 1. n Chapter 11: Sequences and, Section / 40

5 By definition, it means to 1. 1 n(n+1) is convergent and it is equal Very few series have such a beautiful cancellation formula. For most of the series, we cannot compute the partial sum explicitly. Chapter 11: Sequences and, Section / 40

6 Example 3. (Geometric series) Compute n=0 a r n. For what values of r, this series converges. Remark: Recall {a n } is a geometric sequence if a n+1 a n constant r, for all n. Solution: If r 1, s n = a + ar + ar ar n 1 = a 1 r n When r = 1, s n = a + a + a = na. equals a 1 r. (1) Chapter 11: Sequences and, Section / 40

7 How to prove identity (1)? s n = a + ar + ar ar n 1 r s n = ar + ar 2 + ar ar n Subtract the first line by the second line gives (1 r)s n = a ar n which is equivalent to for r 1. s n = a 1 r n 1 r Chapter 11: Sequences and, Section / 40

8 If r < 1, then lim n r n = 0. Thus lim s n = n a 1 r. Chapter 11: Sequences and, Section / 40

9 If r = 1, then the partial sum s n is not equal to a 1 r n 1 r. It is s n = na, whose limit is infinity. If r 1 or r > 1, then limit of r n does not exist. (Recall the four cases in Chapter 11.1, in Oct 29 notes.) Hence limit of s n does not exist. Chapter 11: Sequences and, Section / 40

10 Conclusion: If 1 < r < 1, then the series converges and a + ar + ar 2 + = n=0 If r 1 or r 1, then n=0 ar n diverges. ar n = a 1 1 r. Chapter 11: Sequences and, Section / 40

11 Example 4. Find the sum of geometric series Solution: Note a n+1 a n is a constant for all n, and they all equal to 2 3. Thus it is a geometric series. a = 5, r = a n+1 a n = a 2 a 1 = 2 3. Apparently, r < 1. Thus the series equals to a 1 1 r = ( 2 3 ) = = 3. Chapter 11: Sequences and, Section / 40

12 Example 5. Is the series 52n 2 1 n convergent or divergent? Solution: 5 2n 2 1 n = (5 2 ) n n = 2 ( 25 2 )n. Note r = 25 2 > 1, thus the geometric series diverges. Chapter 11: Sequences and, Section / 40

13 Example 6. Find 00 2 n. Solution: 00 2 n =( 1 2 )100 + ( 1 2 )101 + ( 1 2 )102 + ( 1 2 )103 + =( 1 (1 2 ) (1 2 )2 + ( 1 ) 2 )3 + (2) =( 1 2 )100 n=0 1 2 n. r = 1 2 < 1, thus n=0 1 2 n = = 2. Chapter 11: Sequences and, Section / 40

14 Hence 2 n = ( ) = ( 1 2 ) This example shows that we can compute geometric series starting with any index. Chapter 11: Sequences and, Section / 40

15 Theorem If a n is convergent, then lim n a n = 0. Remark: The converse is not true in general. If lim n a n = 0, the series a n may be convergent and divergent. We will give examples in future. Chapter 11: Sequences and, Section / 40

16 Contrapositive Statement of the Theorem: If lim n a n 0, then a n is divergent. This is also called Divergence test. Example 7. n Solution: Since lim n 2n+4 diverges. n 2n+4 = 1 2 statement of the Theorem, 0, by the contrapositive n 2n+4 diverges. Chapter 11: Sequences and, Section / 40