1. Arithmetic Series Additive Recursion

2. 7/15/2013 Arithmetic Series 2 The art of asking the right questions in mathematics is more important than the art of solving them − Georg Cantor

3. 7/15/2013 Arithmetic Series 3 What is a series ? A sum of a sequence of numbers Definition: A finite series is a sum of form a 1 + a 2 + a 3 +... + a n for some positive integer n NOTE: A finite series has last term a n Series

4. 7/15/2013 Arithmetic Series 4 What is a series ? A sum of a sequence of numbers Definition: An infinite series is a sum of form a 1 + a 2 + a 3 + + a n + NOTE: An infinite series has no last term Series

5. 7/15/2013 Arithmetic Series 5 n th Partial Sum S n S n = a 1 + a 2 + a 3 +... + a n Finite Series Sum S of all n terms is S n for some positive integer n From the closure axiom, S n exists and is a number Partial Sums

6. 7/15/2013 Arithmetic Series 6 Finite Series Sum of n terms can be written as Examples: Summation Notation ∑ akak k=1 n = 2(1) + 2(2) + 2(3) + 2(4) + 2(5) = 30 = 6 + 6 = 24 a 1 + a 2 + a 3 + + a n = k=1 5 2k ∑ k=1 4 ∑ 6 1. 2.

7. 7/15/2013 Arithmetic Series 7 Partial Sums Definition: such that A series ∑ akak k=1 n for some constant d and for all k, k = 1, 2, 3, …, n is an arithmetic series with common difference d a k+1 = a k + d

8. 7/15/2013 Arithmetic Series 8 Arithmetic Sums Partial Sums = a 1 + ( a 1 + d) + ( a 1 + 2d) + + ( a 1 + (n – 2)d) + ( a 1 + (n – 1)d) S n = a 1 + a 2 + a 3 + + a n–1 + a n For arithmetic series ∑ akak k=1 n = S n with common difference d In reverse order … S n = ( a 1 + (n – 1)d) + ( a 1 + (n – 2)d) + + ( a 1 + 2d) + ( a 1 + d) + a 1 SnSn

9. 7/15/2013 Arithmetic Series 9 Arithmetic Sums Partial Sums + ( a 1 + (n – 2)d) + ( a 1 + (n – 1)d) Adding … S n = ( a 1 + (n – 1)d) + ( a 1 + (n – 2)d) + + ( a 1 + 2d) + ( a 1 + d) + a 1 = a 1 + ( a 1 + d) + ( a 1 + 2d) SnSn and … = [ 2 a 1 + (n – 1)d ] + [ 2 a 1 + (n – 1)d ] + 2S n n terms

10. 7/15/2013 Arithmetic Series 10 Arithmetic Sums Partial Sums = [ 2 a 1 + (n – 1)d ] + [ 2 a 1 + (n – 1)d ] + 2S n = n [ 2 a 1 + (n – 1)d ] = n [ a 1 + a 1 + (n – 1)d ] anan = n ( a 1 + a n ) Thus S n = n ( a 1 + a n ) 2 = n (average of a 1 and a n )

11. 7/15/2013 Arithmetic Series 11 a k = area of k th rectangle of unit width S n = area of 1 st n rectangles Let n = 5 Then Arithmetic Series Partial Sums 12345 a1a1 a2a2 a4a4 a5a5 Average height S n = a1 + ana1 + an 2 ( ) n = 5 a1 + a5a1 + a5 2 ( ) S5S5 5 a3a3 a1 + a5a1 + a5 2

12. 7/15/2013 Arithmetic Series 12 Arithmetic Series Infinite Series ( 2 a 1 + (n – 1)d ) = 2 n = n a 1 + n(n – 1) 2 d Recall that S n n a1 + ana1 + an 2 ( ) = What happens to S n as n ∞ ? What can we say about S ∑ k=1 ∞ akak = ? SnSn

13. 7/15/2013 Arithmetic Series 13 Arithmetic Series Infinite Series Remember: The sum of an infinite arithmetic series never exists ! S does not exist ! What can we say about S ∑ k=1 ∞ akak = ? = n a 1 + n(n – 1) 2 d SnSn

14. 7/15/2013 Arithmetic Series 14 Example: Let a 1 = -6 with common difference 4 Arithmetic Series ∞ as n ∞ a n = a 1 + (n – 1)d = -6 + 4(n – 1) a 2 = a 1 + 4 = -6 + 4 = -2 a 3 = a 2 + 4 = -2 + 4 = 2 a 4 = a 3 + 4 = 2 + 4 = 6 = 2n(n – 4 ) = SnSn a1 + ana1 + an 2 ( ) n Thus S does not exist ! = k=1 ∞ akak ∑

15. 7/15/2013 Arithmetic Series 15 Think about it !