HW#7 - ²> 0 there is N = N ² with ν ≤ s N< ν ²...

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SOLUTIONS TO HOMEWORK # 7 10.9 (a) s 1 = 1 s 2 = ± 1 2 (1) 2 = 1 2 s 3 = ± 2 3 ¶± 1 2 2 = 1 6 s 4 = ± 3 4 ¶± 1 6 2 = 1 48 . (b) s 1 = 1 and s n +1 = n n + 1 ( s n ) 2 . We have s 2 n > 0 and n n + 1 > 0, so s n +1 > 0 for all n . We will use induction to prove that s n 1 n N . Indeed: this is true for n = 1; assume s n 1 for some n ; then s 2 n s n and (since n n + 1 < 1) s n +1 = n n + 1 ( s n ) 2 < s n ; (1) hence s n +1 < s n 1 and we obtain by induction s n 1 n N . Now we can see that equation (1) is valid n N , that is { s n } is decreasing. Also 0 < s n < s 1 = 1 so ( s n ) is bounded. Hence by theorem 10.2, it converges. Thus lim s n exists. (c) { s n +1 } differs from { s n } only by the first term so lim s n +1 = lim s n = s . Thus lim s n +1 = lim ± n n + 1 (lim s n ) 2 , s = s 2 , so s = 0 or s = 1. Since s n < 1 2 for all n > 2 , , we have: s 1 2 < 1. Hence s = lim s n = 0. 10.2 Let { s n } be bounded and non-increasing. Since { s n } is bounded, ν := inf { s n } ∈ R . The Approximation Property of the infimum guarantees that for each
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Unformatted text preview: ² > 0 there is N = N ( ² ) with ν ≤ s N < ν + ² . Since s n is non-increasing, ν ≤ s n < ν + ² for all n > N . Thus | s n-ν | < ² for all n > N . Hence lim s n = ν by definition. In other words, { s n } converges to inf { s n } . 1 Problem A . (1) lim 3 n 2 + 6 2 n 2 + 5 n + 1 = lim 3 + 6 n 2 2 + 5 n + 1 n 2 = lim(3 + 6 n 2 ) lim(2 + 5 n + 1 n 2 ) = lim3 + lim 6 n 2 lim2 + lim 5 n + lim 1 n 2 = 3 2 . The same way (2) lim 3 n 3 + 6 2 n 2 + 5 n + 1 = lim 3 n + 6 n 2 2 + 5 n + 1 n 2 = ∞ 2 = ∞ . (3) lim 3 n 3 + 6 2 n 4 + 5 n + 1 = lim 3 n + 6 n 4 2 + 5 n 3 + 1 n 4 = 2 = 0. 2...
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This note was uploaded on 04/28/2008 for the course MATH 312 taught by Professor Olsavskygregorm during the Spring '08 term at Penn State.

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HW#7 - ²> 0 there is N = N ² with ν ≤ s N< ν ²...

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