homework202 - Exercise 1.57(1 The sequence diverges to 1 there is no convergent subsequence(2 The even subsequence converges to 2 and the odd

homework202 - Exercise 1.57(1 The sequence diverges to 1...

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E xercise 1.57 (1) The sequence diverges to -1 , there is no convergent subsequence. (2) The even subsequence converges to 2 and the odd subsequence converges to - 2. By the argument similar to Example 2.4.1, the limits of all convergent subsequences are 2 and - 2. (3) Since ( - 1) n 2 n ( n + 1) ( p n + 2) 3 > 2 n 2 (2 p n ) 3 > p n 4 and lim p n 4 = 1 , we see that lim ( - 1) n 2 n ( n + 1) ( p n + 2) 3 = 1 . Consequently, all the subsequences will diverge to 1 (see Exercise 1.31), and no subsequence converges. (4) The even subsequence converges to 1 and the odd subsequence converges to - 1. By the argument similar to Example 1.5.3, the limits of all convergent subsequences are 1 and - 1. (5) The sequence is a combination of six sequences. x 6 k = 0 , x 6 k +1 = p 3(6 k + 1) 4 , x 6 k +2 = p 3(3 k + 1) 2(( - 1) k +1 (3 k + 1) + 1) , x 6 k +3 = 0 , x 6 k +4 = - p 3(3 k + 2) 2(( - 1) k +1 (3 k + 2) + 1) , x 6 k +5 = - p 3(6 k + 5) 4 . Both x 6 k and x 6 k +3 converge to 0. Both x 6 k +1 and x 6 k ++5 diverge to 1 . Both x 6 k +2 and x 6 k +4 are made up of two subsequences with limits p 3 2 and - p 3 2 . Then by the argument similar to Example 1.5.3, all the limits of convergent subsequences are 0, p 3 2 and - p 3 2 . (6) The even subsequence converges to 0 and the odd subsequence converges to 1. The limits of all convergent subsequences are 0 and 1. (7) The even subsequence converges to 1 and the odd subsequence diverges to + 1 . The only limit of all convergent subsequences is 1. (8) 2 and 3. E xercise 1.58 Any l 2 LIM { x n } is the limit of a subsequence of x n . This subsequence is also a subsequence of z n . Therefore l 2 LIM { z n } . This shows LIM { x n } LIM { z n } . By the same reason, LIM { y n } LIM { z n } . On the other hand, any l 2 LIM { z n } is the limit of a subsequence z n k of z n . By the con- struction of z n , the subsequence z n k either contains infinitely many terms from x n , or contains infinitely many terms from y n . If z n k contains infinitely many terms from x n , then by picking those terms of z n k that are from x n , we get a sequence z n kp that is a subsequence of z n k and also a subsequence of x n . Therefore l = lim k !1 z n k = lim p !1 z n kp 2 LIM { x n } . The 2 LIM { x n } part is due to z n kp being a subsequence of x n . If z n k contains infinitely many terms from y n , then we get l 2 LIM { y n } . Therefore LIM { z n } LIM { x n } [ LIM { y n } . E xercise 1.59
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By lim l k = l , for any > 0 and N , there is k > N , such that | l k - l | < . By l k = lim x n p for some subsequence x n p , there is p > N , such that | x n p - l k | < . In particular, we proved that there is m = n p p > N , such that | x m - l | | x n p - l k | + | l k - l | < 2 . By Proposition 1.5.3, what we proved shows that l is the limit of a convergent subsequence. E xercise 1.60 We only need to find the maxima and minima of the limits found in Exercise 1.57. (1) lim = lim 3 p n = -1 . (2) lim ( - 1) n 2 n + 1 n + 2 = max { - 2 , 2 } = 2, lim ( - 1) n 2 n + 1 n + 2 = min { - 2 , 2 } = - 2. (3) lim ( - 1) n 2 n ( n + 1) ( p n + 2) 3 = + 1 , lim ( - 1) n 2 n ( n + 1) ( p n + 2) 3 = -1 .
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  • Math, Limits, Limit, lim, Limit of a function, Limit superior and limit inferior, subsequence

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