solHMK4 - Section 3.1 #4) Let > 0 be arbitrary....

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Unformatted text preview: Section 3.1 #4) Let > 0 be arbitrary. Select K N such that K > | b | . Then if n K , we have b n = | b | n | b | K < . Therefore, lim n b n = 0. #6) (a) Let > 0 be arbitrary. Select K N large enough so that K > 1 / 2 . Then if n K , we have n + 7 > n K > 1 / 2 . It follows that if n K , then | 1 / n + 7 | < . Therefore, lim n 1 / n + 7 = 0. (b) 2 n n +2- 2 =- 4 n +2 = 4 n +2 < 4 n . Since 1 /n 0, we can apply Theorem 3.1.10 (with C = 4) to get the desired result. (c) n n +1 = n n +1 < n +1 n +1 = 1 n +1 . The limit of 1 n +1 is 0, (proved just like in part (a) of this problem), and so we can apply Theorem 3.1.10 (with C = 1) to get the desired result. (d) (- 1) n n n 2 +1 = n n 2 +1 < n n 2 = 1 n . Since 1 /n 0, we can apply Theorem 3.1.10 (with C = 1) to get the desired result. #8) lim n x n = 0 ( > 0)( K N )( n K ) | x n | < ( > 0)( K N )( n K ) || x n || < lim n | x n | = 0 . Define x n := (- 1) n . Then | x n | = 1 for all n , so lim | x n | = 1 even though ( x n ) has no limit. #13) < b < 1, so 1 /b > 1. Let a = (1 /b )- 1 > 0 so that b = 1 1+ a . (1 + a ) n > 1 2 n ( n- 1) a 2 as in Example 3.1.11(d). So | nb n | = n (1+ a ) n < 2 ( n- 1) a 2 . We can now apply Theorem 3.1.10 since 1 n- 1 0 (we take C = 2 a 2 ). #15) X := ( x n = n 2 n ! : n N } . We prove that X 2 converges to 0. By Theorem 3.1.9, this shows that X converges to 0. For n 2, x n +1 x n = ( n +1) 2 ( n +1)!...
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This note was uploaded on 09/30/2009 for the course MATH 444 taught by Professor Junge during the Spring '08 term at University of Illinois at Urbana–Champaign.

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solHMK4 - Section 3.1 #4) Let > 0 be arbitrary....

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