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Unformatted text preview: ANALYSIS EXERCISE 7SOLUTIONS 1. (a) Since ( n N + ) ( a n = a ), one has evidently that ( > 0)( N N )( n N )[( n > N ) (  a n a  = 0 < )] (Just take N = 1). (b) By the definition ( M > 0)( N N )( n N )[( n > N ) ( a n > M )] . Note that ( a n > M ) ( 1 a n < 1 M ). Fix > 0. Take M = 1 in the above. Then ( n N )[( n > N ) (0 < 1 a n < )] . 2. (a) ( > 0)( N N )( n N )[( n > N ) (  a n a  )] . (b) 2 n n + 1 1 = n 1 n + 1 . Take = 1 2 . If n > 3 the n 1 n +1 > 1 2 . 3. (a) Note that <  a n  1 n 0 ( n ) . By the sandwich rule lim n a n = 0. (b) < 2 n n 3 + 1 < 2 n 0 ( n ) . (c) < 1 n ! < 1 n 0 ( n ) . (d) o < n + 1 n = 1 n + 1 + n < 1 n . Fox > 0. Take N > 1 2 (possible by the Archimedian princi ple). Then ( n > N ) ( 1 n < ) . The required follows from the sandwich rule. 1 2 ANALYSIS EXERCISE 7SOLUTIONS 4. (a) Fix > 0. Take N N such that ( n N ) [( n > N ) (  a n a  < / 2) ....
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This note was uploaded on 02/25/2010 for the course MATHEMATIC 11007 taught by Professor Spyros during the Winter '10 term at Bristol Community College.
 Winter '10
 Spyros

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