chapter2_solution - 2.1.0 Decide which of the following...

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2.1.0 Decide which of the following statements are true and which are false. Prove the true ones and provide a counterexample for the false ones. (a) If x n converges, then x n n also converges. (b) If x n does not converge, then x n n does not converge. (c) If x n converges and y n is bounded, then x n y n converges. (d) If x n converges to zero and y n > 0 for all n N , then x n y n converges.
Proof. (a) True. If x n is convergent, then x n is bounded, say | x n | < M for all n . Then x n n < M n 0 as n → ∞ . Hence, x n n is convergent. (b) False. Let x n = n . Then x n does not converge. But x n n = 1 n 0 as n → ∞ . (c) False. Let x n = 1 and y n = ( - 1) n . Then x n is convergent and y n is bounded. But x n y n = ( - 1) n does not converge. (d) False. Let x n = 1 n and y n = n 2 . Then x n converges to zero and y n > 0 for all n N . But x n y n = n does not converge. 2.1.2 Suppose that x n is a sequence of real numbers that converges to 1 as n → ∞ . Using Definition 2.1, prove that the following limits exists. 1
(b) πx n - 2 x n π - 2 as n → ∞ . Proof. (b) Since lim n →∞ x n = 1, we find that (i) for all ε > 0 there exists a N 1 N such that, for all n > N 1 , | x n - 1 | < ε 4 . (ii) there exists a N 2 N such that, for all n > N 2 , | x n - 1 | < 1 / 2 . This implies that for all n > N 2 , 1 / 2 < x n < 3 / 2 . Hence, for all n > max { N 1 , N 2 } , we find πx n - 2 x n - ( π - 2) = 2( x n - 1) x n < 2 × ε 4 1 / 2 = ε. Hence, lim n →∞ πx n - 2 x n = π - 2. 2.2.0 Determine which of the following statements are true and which are false. Prove the true ones and provide counterexamples for the false ones. (a) If x n → ∞ and y n → -∞ , then x n + y n 0 as n → ∞ . (b) If x n → -∞ , then 1 /x n 0 as n → ∞ . (c) If x n 0, then 1 /x n → ∞ as n → ∞ . (d) If x n → ∞ , then (1 / 2) x n 0 as n → ∞ . Proof. (a) False. Let x n = 2 n and y n = - n . Then lim n →∞ x n = and lim n →∞ y n = -∞ . Moreover, lim n →∞ ( x n + y n ) = lim n →∞ n = . 2
(b) True. If x n → -∞ , then, for all M N , there exists a N N such that, for all n > N , x n < - M . Hence, | x n | > M for all n > N. Now, let ε = 1 M . Then for all n > N , 1 | x n | < 1 M = ε. This implies lim n →∞ 1 x n = 0. (c) False. Let x n = ( - 1) n n . Then lim n →∞ x n = 0. But lim n →∞ 1 x n = lim n →∞ n ( - 1) n does not exist. (d) True. Let a n = ln(1 / 2) x n = x n ln(1 / 2) = - x n ln 2. Since lim n →∞ x n = , we find lim n →∞ a n = -∞ . Hence 0 = lim n →∞ e a n = lim n →∞ e ln(1 / 2) xn = lim n →∞ (1 / 2) x n . 2.2.3 Find the limit (if it exists) of the following sequence (a) x n = 2 + 3 n - 4 n 2 1 - 2 n + 3 n 2 .
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