By definition 162 we prove our conclusion 2 prove the

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By Definition 16.2, we prove our conclusion. (2). Prove: The sequence ( s n ) defined by s n = cos( ) is divergent. Proof By choosing n = 2 k , we have s 2 k = 1 1 as k → ∞ , and by choosing n = 2 k - 1, we have s 2 k - 1 = - 1 → - 1 as k → ∞ . Therefore, by Theorem 19.4 (truly, Example 19.6), we prove that ( s n ) defined by s n = cos( ) is divergent. (Of course, we can prove the statement based on Example 16.12 as well.) 4. (18 points) Find the following limits. (1). lim n →∞ ( n 2 + 2 n - n 2 - 3 n ) Solution. lim n →∞ ( n 2 + 2 n - n 2 - 3 n ) = lim n →∞ ( n 2 + 2 n - n 2 - 3 n )( n 2 + 2 n + n 2 - 3 n ) n 2 + 2 n + n 2 - 3 n = lim n →∞ 5 n n 2 + 2 n + n 2 - 3 n = 5 2 . 2
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(2). lim n →∞ 2 n 2 - 3 1 - 2 n Solution. Since the degree of the numerator is bigger than the degree of the denomina- tor, and the product of the leading coefficients is negative, we have lim n →∞ 2 n 2 - 3 1 - 2 n = -∞ . (3). lim n →∞ n 2 1 . 01 n Solution. Since s n +1 s n = ( n +1) 2 1 . 01 n +1 n 2 1 . 01 n = ( n + 1) 2 1 . 01 n 2 1 1 . 01 < 1 as n → ∞ , by Theorem 17.7, we have lim n →∞ n 2 1 . 01 n = 0. 5. (18 points) Let ( s n ) be defined by s 1 = 2, s n +1 = 2 + s n for n N . (1). Prove that ( s n ) is an irrational sequence, that is, for every n N , s n is an irrational number.
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  • Spring '12
  • Ravindran
  • Topology, Mathematical analysis, Metric space, Cauchy sequence, False. Example

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