4 Let s n be bounded Then there exists a subsequence s n k such that s n k lim

4 let s n be bounded then there exists a subsequence

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4. Let ( s n ) be bounded. Then there exists a subsequence ( s n k ) such that s n k lim sup s n , and there exists a subsequence ( s n l ) such that s n l lim inf s n . This can be done by the definitions of sup ( S ) and inf ( S ) (see Theorem 4.4.11). 82
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5. If ( s n ) is unbounded above, we define lim sup ( s n ) = + . If ( s n ) is unbounded below, we define lim inf ( s n ) = -∞ . 6. Let s n = 1+( - 1) n +1 2 and ( s n ) = (1 , 0 , 1 , 0 , 1 , 0 , ... ). Then 0 and 1 are subsequencial limits, lim sup s n = 1 and lim inf s n = 0. 7. There are many important applications on lim sup s n or lim inf s n . For examples, the formula for the radius of convergence for a power series h =0 a n x n is given by: r = 1 lim sup n p | a n | . 8. Let s n = ( - 1) n n +1 n and ( s n ) = ( - 1 2 , 3 2 , - 4 3 , 5 4 , - 6 5 , ... ). Then - 1 and 1 are sunsequencial limits, lim sup s n = 1, and lim inf s n = - 1. 9. Let s n = - 2 + 1 n , n = 3 k - 2 , 1 n , n = 3 k - 1 , 1 + 1 n , n = 3 k. Then - 2 , 0 and 1 are subsequenctial limits, lim sup s n = 1, and lim inf s n = - 2. 10. Let s n = cos 3 , and ( s n ) = 1 2 , - 1 2 , - 1 , - 1 2 , 1 2 , 1 , ... . Then 1 , 1 2 , - 1 2 , - 1 are subsequenctial limits, lim sup s n = 1, and lim inf s n = - 1. 11. Let s n = n sin 2 2 . Then lim sup s n = , and lim inf s n = 0. 12. Let s n = n 2 cos nπ = ( - 1) n n 2 . Then lim sup s n = , and lim inf s n = -∞ . 13. Let s n = - 2 + 1 n , n = 3 k - 2 , 1 n , n = 3 k - 1 , 1 + 1 n , n = 3 k, n 1 . Then lim sup s n = 1, sup( s n ) = 4 3 , lim inf s n = - 2 and inf( s n ) = - 2. 83
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14. Let s n = sin 6 . Then lim sup s n = 1, sup( s n ) = 1, lim inf s n = - 1, and inf( s n ) = - 1. 15. Let s n = sin 6 + 1 n . Then lim sup s n = 1, sup( s n ) = 3 2 , lim inf s n = - 1, and inf( s n ) = - 1. Theorem 0.5 The limit lim(1 + 1 n ) n exists. Proof: We want to show s n = 1 + 1 n n is increasing sequence. In fact, by the inequality of arithmetic and geometric means: Let a 1 , ..., a n be positive real numbers. Then a 1 + a 2 + ... + a n n n a 1 a 2 ...a n , i.e., a 1 + a 2 + ... + a n n n a 1 a 2 ...a n . In particular, if a 1 = a 2 = ... = a n = x and a n +1 = y , we have nx + y n + 1 n +1 x n y. (2) Let x = 1 + 1 n and y = 1. Then from (2), we find n (1 + 1 n ) + 1 n + 1 n +1 1 + 1 n n . i.e., s n +1 = 1 + 1 n + 1 n +1 1 + 1 n n = s n . Also we let t n = 1 + 1 n n +1 . We claim that ( t n ) is decreasing. In fact, taking x = 1 - 1 n and y = 1 in (2), we get n (1 - 1 n ) + 1 n + 1 n +1 1 - 1 n n , 84
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i.e., n n + 1 n +1 n - 1 n n , which implies n n - 1 n n + 1 n n +1 , i.e., 1 + 1 n - 1 ) n 1 + 1 n n +1 , i.e., t n - 1 t n for all n . We also notice s n t n so that s 1 s 2 s 3 ... t 3 t 2 t 1 .
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