1. A sequence can be thought of as a list of numbers written in a definite order: The number a1 is called the first term, a2 is the second term, and in general an is the nth term. A function whose domain is the set of positive integers a1, a2, a3, a4, ,a n,

2. NOTATION The sequence {a1 ,a2 ,a3 , . . .} is also denoted by {an} or =1}{ nna 3. In the following examples, we give three descriptions of the sequence: 1. Using the preceding notation 2. Using the defining formula 3. Writing out the terms of the sequence Example 1 4. Preceding Notation Defining Formula Terms of Sequence ( 1) ( 1) 3 n n n + ( 1) ( 1) 3 n n n n a + = 2 3 4 5 , , , ,..., 3 9 27 81 ( 1) ( 1) ,... 3 n n n + ( 1) ( 1) 3 n n n n a + =( 1) ( 1) 3 n n n + 2 3 4 5 , , , ,..., 3 9 27 81 ( 1) ( 1) ,... 3 n n n + { } 3 3 n n = 3 ( 3) na n n = { }0,1, 2, 3,..., 3,...n 5. CONVERGENT AND DIVERGENT SEQUENCE CONVERGENT---- The sequence (An) in R is said to converge if there exists a number L for all > 0, there exists a natural number N such that L < an < L+ for all n N.The sequence has a limit , we can say that the sequence is convergent DIVERGENT---- If a sequence has no limit , we can say that the sequence is called divergent sequence 6. Example Ex. Is the series convergent or divergent? Sol. Since it is a series with thus is divergent. Ex. Find the sum of the series Sol. 2 1 1 2 3n n n = 4/3 1r = > 2 1 1 1 1 1 1 4 2 3 4 3 4( ) 3 n n n n n n n n = = = = = 1 1 1 . 1 3 2 4 ( 2)n n + + + + + L L 1 1 1 1 1 1 1 1 ( ) (1 ) 2 2 2 2 1 2 n n k s k k n n= = = + + + + 3 lim 4 n n s s = = 7. Example Ex. Show that the series is divergent? Sol. 1 1 1 1 1 1 2 3 4n n = = + + + + L 1 2 1 1 1 2 1 1, 1 , 2 3 4 4 2 s s= = + + > = 1 1 1 1 4 1 1 1 8 1 , 5 6 7 8 8 2 9 16 16 2 + + + > = + + > =L 2 1 2 n n s > + 8. Example Ex. Determine whether the series converges or diverges. Sol. The improper integral So the series diverges. 2 1 1 ln ln 2 x x dx x = = 1 ln n n n = 1 ln n n n = 9. Test for convergence or divergence of: 2 3 n na n = 1 2 3 n n n = 1 1 2 ( 1) 3 n n na + + + = 1 1 2 ( 1) 2 3 3 n n n n n a a n + + + = 1 1 2 3 n n n n + + = 1 2 3 n n + = 1 lim n n n a a + 2 1 lim 3 n n n + = 2 3 = Since this ratio is less than 1, the series converges. 10. Test for convergence or divergence of: 2 2 n n n a = 2 11 ( 1) 2nn n a ++ + = 2 2 1 1 ( 1) 2 2 n n n n na n a + + + = ( ) 2 1 2 1 2 2 n n n n+ + = 2 2 1 ( 1) 2 2 n n n n + + = 1 lim n n n a a + Since this ratio is less than 1, the series converges. 2 1 2n n n = 2 2 1 2 1 2 n n n + + = 2 2 1 2 1 lim 2 n n n n + + = The ratio of the leading coefficients is 1 1 2 = 11. LIMITS OF SEQUENCES A sequence {an} is called: Increasing, if an < an+1 for all n 1, that is, a1 < a2 < a3 < Decreasing, if an > an+1 for all n 1 Monotonic, if it is either increasing or decreasing 12. Example Ex. Find the limit Sol. . 1 2 1 1 1 lim 222 + ++ + + + nnnnn L nn n nnnnnnnn + = + ++ + + ++ + + + 222222 111 2 1 1 1 LL 11 1 1 11 2 1 1 1 222222 + = + ++ + + ++ + + + n n nnnnnn LL 1 1 1 1 lim 1 lim,1 1 1 1 limlim 2 22 = + = + = + = + n n n n nn n nnnn 13. SEQUENCES 14. SEQUENCES 15. DECREASING SEQUENCES Q-Show that the sequence is decreasing Sequence is decresing beacause and so an > an+1 for all n 1. Example 11 3 5n + 3 3 3 5 ( 1) 5 6n n n > = + + + + 16. Q-Show that the sequence is decreasing. 2 1 n n a n = + 2 2 1 ( 1) 1 1 n n n n + < + + + We must show that an+1 < an, that is, The inequality is equivalent to the one we get by cross-multiplication: 2 2 2 2 3 2 3 2 2 1 ( 1)( 1) [( 1) 1] ( 1) 1 1 1 2 2 1 n n n n n n n n n n n n n n n n + < + + < + + + + + + + + < + + < + 17. BOUNDED SEQUENCES A sequence {an} is bounded: Above, if there is a number M such that an M for all n 1 Below, if there is a number m such that m an for all n 1 If it is bounded above and below 18. 1.The sequence an = n is bounded below (an > 0) but not above. 2.The sequence an = n/(n+1) is bounded because 0 < an < 1 for all n. 19. Theorem A convergent sequence of real numbers is bounded Proof: Suppose that lim (xn) = x and let := 1. By Theorem 3.1.6, there is a natural number K := K(1) such that if n K then |xn x| < 1. Hence, by the Triangle Inequality, we infer that if n K, then |xn| < |x| + 1. If we set M := sup {|x1|, |x2|, , |xK-1|, |x| + 1}, then it follows that |xn| M, for all n N.