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Unformatted text preview: Section 3.4 #4) (a) x n = 1 ( 1) n + 1 /n . We take the subsequence of the evenindexed terms: x n k = x 2 k = 1 ( 1) 2 k + 1 / (2 k ) = 2 + 1 / (2 k ). So ( x 2 k ) converges to 2. Next we take the subsequence of oddindexed terms: x 2 k 1 = 1 ( 1) 2 k 1 + 1 / (2 k 1) = 1 / (2 k 1). So ( x 2 k 1 ) converges to 0. Since there are two subsequences converging to different limits, the entire sequence diverges. (b) x n = sin( n/ 4). We first take the subsequence of multiples of 8: x 8 k = sin(8 k/ 4) = sin(2 k ) = 0. So the subsequence ( x 8 k ) converges to 0. Now we take the subsequence where n k = 8 k + 2: x 8 k +2 = sin((8 k + 2) / 4) = sin(2 k + / 2) = sin( / 2) = 1. So ( x 8 k +2 ) converges to 1. Since there are two subsequences converging to different limits, the entire sequence diverges. #6) (a) x n +1 < x n ( n + 1) 1 / ( n +1) < n 1 /n ( n + 1) n < n n +1 ( n + 1) n n n < n 1 + 1 n n < n It was shown in Example 3.3.6 that (1+1 /n ) n < 3. Hence, for n 3, we have (1+1 /n ) n < 3 n , and therefore x n +1 < x n for all n 3. Thus, the sequence is ultimately decreasing (its 3tail is decreasing) and since it is bounded below by 1, x := lim( x n ) exists. Also x 1. (b) ( x 2 n ) also converges to x , and x 2 n = (2 n ) 1 / (2 n ) = 2 1 /n n 1 /n = 2 1 /n x n . Taking the limit of both sides and recalling that 2 1 /n converges to 1, we arrive at x = x . Since x 1, the only viable solution is x = 1. #7) (a) (1 + 1 /n 2 ) n 2 is a subsequence of ((1 + 1 /N ) N ), which we know converges to e . Therefore, the sequence in question also converges to e . (b) ( (1 + 1 / 2 n ) 2 n ) is a subsequence of ((1 + 1 /N ) N ), which we know converges to e . Therefore, the subsequence also converges to e . Hence, ((1 + 1 / 2 n ) n ) = p (1 + 1 / 2 n ) 2 n e . (c) (1 + 1 /n 2 ) 2 n 2 = (1 + 1 /n 2 ) n 2 2 . So the sequence in question converges to e 2 by part (a) and the algebraic properties of limits. (d) There is no way to manipulate (1+2 /n ) n to get a subsequence of (1+1 /N ) N . By problem 5 (proven below), it suffices to show that the evenindexed and oddindexed subsequences (1 + 2 / (2 k )) 2 k and (1 + 2 / (2 k 1)) 2 k 1 converge to the same thing. Evenindexed sequence: (1 + 2converge to the same thing....
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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.
 Spring '08
 JUNGE

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