sol4 - SOLUTIONS FOR HOMEWORK 4 3.1.5. (b) Pick ε > 0,...

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Unformatted text preview: SOLUTIONS FOR HOMEWORK 4 3.1.5. (b) Pick ε > 0, and show that there exists N ∈ N s.t. |2n/(n + 1) − 2| < ε for n > N . A simple calculation shows that |2n/(n + 1) − 2| = 2/(n + 1). We know that there exists N ∈ N s.t. 1/N < ε/2. Then |2n/(n + 1) − 2| < ε for n > N . 3.1.10. Let ε = x/2. Then there exists M ∈ N s.t. |x − xn | > ε for any n > M . By the triangle inequality, for such n x x xn ≥ x − |x − xn | > x − = > 0. 2 2 3.1.14. We show that lim(2n)1/n = 1. Clearly, (2n)1/n > 1. Fix ε > 0, and show that (2n)1/n < 1 + ε for n > N (N ∈ N depends on ε). Indeed, (2n)1/n < 1 + ε iff n 2n < (1 + ε)n = k =0 2 nk ε. k Find N ∈ N s.t. N > 4/ε + 1, or in other words, 2N < N (N − 1)ε2 /2. Then, for any n > N , n2 ε < (1 + ε)n , 2n < 2 as desired. 3.1.16. For n ≥ 3, hence, for such n, n! = 1 · 2 · 3 · 4 . . . · n ≥ 1 · 2 · 3 · 3 . . . · 3 = 2 · 3n−2 , 2n 2 ≤2 n! 3 n−2 = (here, we use the techniques of Example (b), p. 58 of the textbook). For any ε > 0 find N ∈ N s.t. N ≥ 3, and 1/N < ε/4. Then 0 < 2n /n! < ε for any n > N . 3.2.7. In Theorem 3.2.3, both sequences are assumed to converge, which need not happen in our case. Thus, that theorem is inapplicable here. Find A > 0 s.t. |bn | < A for any n. Fix ε > 0, and show that there exists N ∈ N s.t. |an bn | < ε for any n > N . There exists N ∈ N s.t. |an | < ε/M for n > N . For such n, |an bn | < ε, and we are done. 3.2.9. Multiplying by the “conjugate,” we get: √ √ √ √ 1 n+1+ n yn = ( n + 1 − n) · √ √ =√ √, n+1+ n n+1+ n √ hence 0 < yn < 1/(2 n). For any ε > 0, find N ∈ N s.t. 1/N < 4ε2 . Then |yn | < ε for any n > N , hence limn yn = 0. 1 2 2 4 < = n (1 + 1/2) n · 1/2 n 2 SOLUTIONS FOR HOMEWORK 4 √ ′ ′ Let yn = nyn , and xn = 1/yn = 1+ 1 + 1/n. By Theorem 3.2.10, lim 1 + 1/n = ′ 1. Therefore, by Theorem 3.2.3, lim xn = 2, and furthermore, lim yn = 1/ lim xn = 1/2. 3.2.12. Let xn = = But lim (1 + a/n)(1 + b/n) = lim(1 + a/n)(1 + b/n) = 1, hence lim (1 + a/n)(1 + b/n) + 1 = 2. Furthermore, lim(a + b + ab/n) = a + b, thus lim xn = (a + b)/2. 3.2.16. (a) Let xn = 1 for each n. Then xn+1 /xn = 1 for each n, and lim xn = 1. (b) Let xn = n. Then xn+1 /xn = 1 for each n, and the sequence (xn ) is unbounded, hence divergent. 3.2.18. (d) lim(n!/nn ) = 0. To establish this, let k = ⌊n/2⌋, and note that 0≤ (n + a)(n + b) − n = ( n(a + b) + ab n (1 + a/n)(1 + b/n) + n (n + a)(n + b) − n) · = a + b + ab/n (1 + a/n)(1 + b/n) + 1 (n + a)(n + b) + n (n + a)(n + b) + n . 1 · . . . · k (k + 1) · . . . · n kk n! n = · ≤ ≤ 2−k ≤ 21−n/2 = 2 2−1/2 . n k n−k n n n n −1/2 n By Example 3.1.11(b), lim(2 ) = 0, hence, by Squeeze Theorem, lim(n!/nn ) = 0. 3.2.19. Fix r ∈ (L, 1), and set ε = r − L. Then |xn − L| < ε for n > N , hence 1/n xn < L + ε = r . Therefore, 0 ≤ xn ≤ r n . An application of Squeeze Theorem and Example 3.1.11(b) shows that lim xn = 0. 3.3.2. We know that x1 > 1, and xn+1 = 2 − 1/xn . We use induction to show that this sequence is decreasing. More precisely, we show hat 1 < xn+1 < xn for every n. To this end, it suffices to establish the following: if a > 1, then 1 < 2 − 1/a < a. The left hand side inequality is obvious, and the right hand side is equivalent to 1 < (a + 1/a)/2. However, 1 ≤ (a + 1/a)/2 by the Arithmetic-Geometric Mean Inequality. Moreover, the equality holds iff a = 1/a, which is not the case for us. Hence, 2 − 1/a < a. 3.3.4. Assume for a moment that the sequence (xn ) has a limit, call it x. Then √ √ x = lim xn+1 = lim 2 + xn = 2 + x, hence (solving the equation x2 = 2 + x) x = 2. We shall prove that, for each n, xn√ xn+1 < 2. To this end, it suffies to establish the < following: if 0 < √ < 2, then a < a + 2 < 2. √ a However, a + 2 > a + a = 2a > a · a, √ √ hence a + 2 > a2 = a. On the other hand, 2 + a < 2 + 2 = 2. Thus, the sequence (xn ) is increasing, and bounded by 2 from above. As we have already seen, lim xn = 2. 1/n SOLUTIONS FOR HOMEWORK 4 3 3.3.11. Note that the sequence (yn ) is increasing. Indeed, 1 1 1 1 + − = > 0. yn+1 − yn = 2n + 1 2n + 2 n + 1 (2n + 1)(2n + 2) On the other hand, yn is a sum of n terms, each less than 1/n, hence yn < 1 for each n. Therefore, the sequence (yn ) is increasing, and bounded above. Therefore, it converges. Back to: the syllabus, the main page of the course, the assignment. ...
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This note was uploaded on 09/30/2009 for the course MATH 444 taught by Professor Junge during the Spring '08 term at University of Illinois at Urbana–Champaign.

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