# sole15 - ANALYSIS EXERCISE 15SOLUTIONS 1 Set Am:= 1 an A:=...

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ANALYSIS EXERCISE 15–SOLUTIONS 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 ( α a n + β b n ). Then C m = α A m + β 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 α A + β 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

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2 ANALYSIS EXERCISE 15–SOLUTIONS (b) Take n = 3 and x = 1. Then e - 1 + 1 1! + 1 2! + 1 3! < 2 4! or e - 8 3 < 1 12 The result follows. 3. The left-hand inequality is clear. We have e x = 1 + x 1! + · · · + x n - 1 ( n - 1)! + X r = n x r r ! Now, X r = n x r r ! = x n n !
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