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Unformatted text preview: HOMEWORK 3: SOLUTIONS/HINTS 8.4 Let > 0, then there is an N such that sn  < M for n N . So we get that sn tn  M sn  < for all n N and hence lim(sn tn ) = 0. 8.5 Let > 0, then we have N N such that an  s < as well as bn  s < for n N. So for n N , we have  < an  s sn  s bn  s < and therefore sn  s < for such n and we get that lim sn = s. 8.7(c) cf. the solution in the book. 8.8(b) Let sn = n2  n  n, then n n2 + n + n sn = sn = 2+n+n 2+n+n n n By the same trick we get that 1 1 n sn  = . 2 + n + n)2 2 2n 2( n
1 1 So let N > 2 and we have that sn  2 < for n N. 1 8.10 Let s := lim sn and let := 2 (s  a) > 0. Then there is an N N such that sn  s < for n N. So a < a + = 1 (s + a) = s  < sn for n N. 2 9.6 (a) Suppose a = lim xn R, then by the limit theorems we get that 3a2 = 3(lim xn )2 = lim 3x2 = lim xn+1 = a. n
1 Hence 3a2 = a and therefore a = 0 or a = 3 . (b) The sequence xn is monotone increasing and unbounded and therefore lim xn = +. (c) In (a) we assumed that a is a real number, when we solved the equation 3a2 = a. But in (b) we saw that this is not the case. Moreover, you can think of + as a further solution of 3a2 = a. So we do not have a contradiciton. 9.9 See the solution in the book. 9.15 Apply exercise 9.12. 9.12 We only do part (a). Let L := lim sn+1 < 1. Then there is a R such sn that L < a < 1. Let := a  L, then there exists N N such that  < sn+1  L < for n N. Therefore sn+1 < L + = a for n N sn sn and we get that sn sn1 sN +1 sn  = sN  < anN sN  sn1 sn2 sN for n N . As a < 1, the right hand side converges to 0 and therefore sn  respectively sn converges to 0.
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 Spring '11
 Peter

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