2 1 Indeed the sequence 0 2 2 2 represents the geometric series 2 i n 2 3 i

2 1 indeed the sequence 0 2 2 2 represents the

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2 1 Indeed, the sequence (0 ,..., 0 , 0 , 2 , 2 , 2 ) represents the geometric series 2 i = n +2 3 - i which converges to 3 - ( n +2) 2 1 - 1 3 = 3 - ( n +1) which is also represented by (0 0 , 1 , 0 , 0 , 0 ). 2 This result is surprising . At the k -th step in constructing C , we removed 2 k - 1 intervals of length 3 - k from [0 , 1], and so the total length removed is k =1 2 k - 1 3 k = 1 3 k =0 2 k 3 k = 1. So, even though we removed almost all of [0 , 1] in the construction, the resulting “dust” C can still be put into 1-1 correspondence with [0 , 1]. 1
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MATH 117 HOMEWORK #6 SOLUTIONS 15.3: (a) First, d 2 (( x 1 ,y 1 ) , ( x 2 2 )) 0 since both | x 1 - x 2 | ≥ 0 and | y 1 - y 2 | ≥ 0. Furthermore, d 2 (( x 1 1 ) , ( x 2 2 )) = 0 if and only if max {| x 1 - x 2 | , | y 1 - y 2 |} = 0, which happens if and only if | x 1 - x 2 | = 0 and | y 1 - y 2 | = 0, or equivalently ( x 1 1 ) = ( x 2 2 ). d 2 is symmetric since taking absolute values is symmetric. Finally, we show d 2 satisfies the triangle inequality. d 2 (( x 1 1 ) , ( x 3 3 )) = max {| x 1 - x 3 | , | y 1 - y 3 |} = max {| x 1 - x 2 + x 2 - x 3 | , | y 1 - y 2 + y 2 - y 3 |} ≤ max {| x 1 - x 2 | + | x 2 - x 3 | , | y 1 - y 2 | + | y 2 - y 3 |} ≤ max {| x 1 - x 2 | , | y 1 - y 2 |} +max {| x 2 - x 3 | , | y 2 - y 3 |} = d 2 (( x 1 1 ) , ( x 2 2 ))+ d 2 (( x 2 2 ) , ( x 3 3 )). Hence, d 2 is a metric on R 2 . (b) - 1 - 1 1 1 Figure 1 16.6: (a) Let ² > 0. Then | ² k | > 0. Let N be sufficiently large so that 1 N < | ² k | . Then for any n > N , | k n | < ² . Hence, lim n →∞ k n = 0. (b) Let ² > 0. Then ² 1 /k > 0. Let N be sufficiently large so that 1 N < ² 1 /k . Then for n > N , 1 n k < 1 N k < ² k/k = ² . Hence, lim n →∞ 1 n k = 0. (c) Let ² > 0 and let N > 1 ² . Then for n > N , 3 n + 1 n + 2 - 3 < 3 n + 1 n - 3 = 1 n < 1 N < ² . Hence, lim n →∞ 3 n +1 n +2 = 3. (d) Let ² > 0 and let N > 1 ² . Then for n > N , sin n n 1 n < 1 N < ² . Hence, lim n →∞ sin n n = 0. (e) Let ² > 0 and let N be sufficiently large so that 1 N - 2 < ² . Then for n > N , n + 2 n 2 - 3 < n + 2 n 2 - 4 = 1 n - 2 < 1 N - 2 < ² . Hence, lim n →∞ n +2 n 2 - 3 = 0. 16.8: (a) The sequence a n = 2 n is unbounded and thus divergent. (b) Let ² = 1 2 and let b R . The ² = 1 2 neighborhood of b may contain either 1 or - 1 but it will not contain both. Hence, for any N N , there will be terms of the sequence b n , with n > N , not contained in N ( b, 1 2 ). Therefore, b n does not converge to any b R which is to say that b n diverges. (c) Let ² = 1 4 and use a similar argument as in part (b). (d) The sequence d n is unbounded and thus divergent. 16.11: Suppose the sequence s n converges to s . Let ² > 0. Since s n s , there is N s N such that for n > N s , | s n - s | < ² . Let N t > N s - k . Then for all n > N t , | t n - s | = | s n + k - s | < ² since if n > N t then n + k > N s . Therefore, t n s . The proof of the converse is similar (let N s > N t + k ). 1
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MATH 117 HOMEWORK #7 SOLUTIONS 17.3: (a) lim n →∞ 3 n 2 + 4 n 7 n 2 - 5 n = lim n →∞ 3 + 4 n 7 - 5 n = lim n →∞ (3 + 4 n ) lim n →∞ (7 - 5 n ) = 3 + 4 lim n →∞ 1 n 7 - 5 lim n →∞ 1 n = 3 7 (b) Similar to (a). Answer is 0.
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  • Fall '08
  • Akhmedov,A
  • Math

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