This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: ANALYSIS EXERCISE 15SOLUTIONS 1. Set A m := m 1 a n , A := 1 a n , B m := m 1 b n , B := 1 b n and C m := m 1 ( a n + b n ). Then C m = A m + B m . Also, A m A as m and B m B as m from the definition of convergence. By properties of convergent sequences (Theorem 1.1.6, (i)), C m A + B as m . This proves the result. 2. (a) We have e x 1 + x 1! + + x n n ! = X r = n +1 x r r ! = x n +1 ( n + 1)! X r = n +1 ( n + 1)! r ! x r n 1 = x n +1 ( n + 1)! X r =0 ( n + 1)! ( n + 1 + r )! x r < x n +1 ( n + 1)! X r =0 ( n + 1)! ( n + 1 + r )! ( n/ 2) r For r 1, ( n + 1)! ( n + 1 + r )! = 1 ( n + 2) ( n + 1 + r ) ( r terms) 1 n r . Continuing from the above, x n +1 ( n + 1)! X r =0 1 n r ( n/ 2) r x n +1 ( n + 1)! X r =0 (1 / 2) r = 2 x n +1 ( n + 1)! . 1 2 ANALYSIS EXERCISE 15SOLUTIONS (b) Take n = 3 and x = 1. Then e 1 + 1 1!...
View Full
Document
 Winter '10
 Spyros

Click to edit the document details