Unformatted text preview: SOLUTIONS FOR HOMEWORK 5 3.3.13. (d) Let xn = (1 − 1/n)n , yn = (1 + 1/n)n , and zn = xn yn = (1 − 1/n2 )n . We know that lim yn = e. Show that lim zn = 1. Indeed, zn < 1, and, by Bernoulli’s Inequality (p. 29 of the textbook), zn ≥ 1 + n · (−1/n2 ) = 1 − 1/n. By Squeeze Theorem, lim zn = 1. Finally, xn = zn /yn , hence lim xn = lim zn / lim yn = 1/e. Remark. For any t ∈ R, et = lim(1 + t/n)n = lim
n k k =0 t /k !. 3.4.3. Let xn = fn+1 /fn , and L = lim xn . For n ≥ 2, (1) xn = fn + fn−1 fn−1 1 fn+1 = = +1=1+ . fn fn fn xn−1 Note that the sequence (fn ) is increasing, hence xn ≥ 1, and, by (1), xn ≤ 2. Thus, 1 ≤ L ≤ 2. Taking the limit of both sides of (1), we obtain: 1 . xn−1 L √ Solving the quadratic equation for L, we obtain L = (1 + 5)/2. L = lim xn = 1 + lim =1+ 1 3.4.10. Note that the sequence (sn ) is decreasing, hence it converges to S = inf sn . We have to ﬁnd a sequence n1 < n2 < . . . s.t. limk xnk = S . To this end, it suﬃces to establish the following: for any ε > 0 and N ∈ N, there exists n > N s.t. xn − S  < ε. Indeed, once the claim has been established, let n1 = 1, and, for k > 1, ﬁnd (inductively) nk > nk−1 s.t. xnk − S  < 2−k . Then limk xnk = S . So, suppose ε > 0 and N ∈ N are given. Then there exists M > N s.t. S ≤ sM < S + ε. By deﬁnition, sM = sup{xM , xM +1 , . . .}, hence there exists n ≥ M s.t. sM − ε < xn ≤ SM . Therefore, S − ε ≤ sM − ε < xn ≤ sM < S + ε, which implies xn − S  < ε. 3.4.11. Let L = lim(−1)n xn , and show that L = 0. Indeed, let yn = (−1)n xn . Then lim y2n = lim y2n−1 = L. However, y2n ≥ 0 for any n, hence L ≥ 0. On the other hand, y2n−1 ≤ 0 for any n, which implies L ≤ 0. Thus, L = 0. Therefore, for any ε > 0 there exists N ∈ N s.t. yn  < ε for n > N . But yn  = xn , hence lim xn = 0. 3.5.6. Let xn =
n k =1 1/k . The, for any p ∈ N, xn+p − xn = 1 p <, k n k =n+1
1
n+p 2 SOLUTIONS FOR HOMEWORK 5 hence limn (xn+p − xn ) = 0 for any p. However, the sequence (xn ) is not Cauchy: for any n ∈ N, 2n 1 n 1 x2n − xn = ≥ =. k 2n 2 k =n+1 3.5.11. For n ≥ 1, let xn = yn+1 − yn . Then, for n ≥ 1, 2 2 2 1 xn+1 = yn + yn−1 − yn = (yn−1 − yn ) = − xn . 3 3 3 3 By Theorem 3.5.8, the sequence (yn ) converges. Furthermore, xn = (−2/3)n−1 x1 for any n, hence
n−1 n−1 yn = y1 +
k =1 xk = y1 + x1
k =1 n−1 − 2 3 k −1 = y1 + x1 As lim(−2/3)n = 0, we conclude that 3 y 2 + 2y 1 3 . lim yn = y1 + (y2 − y1 ) = 5 5 3.5.13. xn+1 = 2 + 1/xn , and x1 = 2, hence xn ≥ 2 for any n. For n ≥ 2 1 xn−1 − xn 1 − = . xn+1 − xn = xn xn−1 xn xn−1 However, xn xn−1 ≥ 4, hence xn+1 − xn  ≤ xn−1 − xn /4, which implies that our sequence is contractive. Denote its limit (which exists by Theorem 3.5.8) by x. Then 1 1 x = lim xn+1 = lim 2 + =2+ , xn x 2 hence x − 2x − 1 = 0. Solving this quadratic equation (and keeping in mind that √ x ≥ 2), we conclude that x = 1 + 2. 3.7.8. Yes. If an converges, then lim an = 0 (this follows from Cauchy’s criterion for convergence). Let bn = a2 . Then lim bn /an = lim an = 0, hence bn converges n by Limit Comparison Test (Theorem 3.7.8). √ 3.7.9. NO. Let an = 1/n2 . Then an converges (Example 3.7.6(c)), while an diverges (it is the harmonic series, described in Example 3.7.6(b)). 3.7.10. YES. Recall the ArithmeticGeometric Mean Inequality: for nonnegative √ √ x and y , xy ≤ (x + y )/2. Let sn = n=1 ak , and tn = n=1 ak ak+1 . Note that k k the sequences (sn ) and (tn ) are increasing. Furthermore,
n 1 − (−2/3) 1 − (−2/3) = y1 + 3x1 1− 5 − 2 3 n−1 . tn ≤ k =1 ak + ak+1 ≤ sn+1 ≤ lim sn , 2 hence it is bounded. As a bounded monotone sequence, (tn ) must converge. SOLUTIONS FOR HOMEWORK 5 3 3.7.11. Note that bn ≥ xn = a1 /n, and xn diverges (apply Limit Comparison Test to xn and 1/n). By Comparison test (3.7.7), bn diverges. Problem A (a bonus problem – very little partial credit is given): Prove that the Euler number e (deﬁned in Example 3.3.6) is irrational. For n ∈ N, consider an = n=0 1/k !. We have proved in class that a1 < a2 < . . . < 3, k hence the sequence (an ) converges. Moreover, e = lim an . Suppose, for the sake of contradiction, that e is rational. As e ≥ a2 > 2, we can write e = M/N , with M, N ∈ N. Increasing N if necessary, we can assume N ≥ 2 (here, N and M need not be mutually prime). Then e = K/N !, where K = M (N − 1)!. Note that aN < e, and we can write aN = KN /N ! for some positive integer N . Then e − aN = (K − KN )/N ! ≥ 1/N !. Now let bn = an − aN . Then lim bn = e − aN ≥ 1/N !. However, for n > N , 1 1 1 1 bn = = ≤ k ! k=N +1 1 · . . . N · (N + 1) · . . . · k N ! k=N +1 N + 1 k =N +1 bn ≤
n n n k −N , hence, by the formula for the sum of the geometric sequence, 1 1 1/(N + 1) − 1/(N + 1)n−N +1 < . N! 1 − 1/(N + 1) N · N! Therefore, lim bn ≤ 1/(N · N !), which contradicts lim bn = e − aN ≥ 1/N !. 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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 Bernoulli

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