1 2 SOLUTIONS FOR HOMEWORK 6 Consequently Theorem 107 lim sup d n lim inf d n 6

1 2 solutions for homework 6 consequently theorem 107

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2 SOLUTIONS FOR HOMEWORK 6 Consequently (Theorem 10.7), lim sup d n = lim inf d n = 6 / 7 (the se- quence converges, hence, by Theorem 9.1, it is bounded). The set of subsequential limits of ( d n ) is { 6 / 7 } . 11.9. (a) We have to show that, for any sequence ( s n ) [ a, b ], conver- gent to s , we have lim s n [ a, b ]. However, by Exercise 8.9 (we proved this in class), a s b , which is what we need. (b) The open interval (0 , 1) is not closed. Consider, for instance, the sequence s n = 1 / ( n +1). Then s n (0 , 1) for any n , yet lim s n / (0 , 1). By Theorem 11.8, (0 , 1) cannot be the set of subsequential limits for any sequence. 12.4. s k sup n>N s n and t k sup n>N t n for any k > N . Thus, s k + t k sup n>N s n + sup n>N t n for any k > N . Taking the supremum of the left hand side over k , we obtain: sup k>N ( t k + s k ) sup n>N s n + sup n>N t n . Therefore, lim sup( t k + s k ) = lim N sup k>N ( t k + s k ) lim ( sup n>N s n + sup n>N t n ) = lim sup n>N s n + lim sup n>N t n = lim sup s n + lim sup t n . 12.5. By Exercise 11.8(a), lim inf a n = lim sup( a n ) for any se- quence ( a n ). By Exercise 12.4, lim sup( t n s n ) lim sup( t n ) + lim sup( s n ). Therefore, lim inf( t n + s n ) = lim sup( t n s n ) ≥ − ( lim sup( t n ) + lim sup( s n ) ) = lim sup( t n ) lim sup( s n ) = lim inf s n + lim inf t n . 12.8. It suffices to show that lim sup s n t n st for any s > lim sup s n and t > lim sup t n . Recall that lim sup s n = lim n v n , where v n = sup k n s k . Therefore, there exists N 1 N s.t. v n < s whenever n N 1 . Therefore, s k < s for k N 1 . Similarly, there exists N 2 N s.t. t k
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  • Fall '08
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  • Limits, lim, Supremum, lim sup, subsequence, lim sup sn

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