Sample Solutions for Midterm I review problems

1 Sample Solutions for Midterm I review problems Math 451 2014 11.3) a) (s6n) = (1) is a monotone subsequence of (sn). (tn) is decreasing so any subsequence of (tn) is decreasing, hence monotone. (u2n) = ( 221n ) is a decreasing subsequence of (un). 1 (v2n) = (1 + 2n ) is a decreasing subsequence of (vn). The answers to (c), (d) and (e) are in the back of the book. We will explain the answers in the case of (sn ). Notice that if N is any natural number then {sn |n > N } = {1, 21 , , −1 2 , −1}, so uN = −1 and vN = 1 for all N . Therefore, lim sup sn = lim vN = lim 1 = 1 and lim inf sn = lim uN = lim −1 = −1. Since lim sup sn 6= lim inf sn , Theorem 10.7 implies that lim sn does not exist, so (sn ) does not converge, diverge to +∞ or diverge to −∞. However, notice that −1 ≤ sn ≤ 1 for all n ∈ N, so (sn ) is bounded. 2 n 14.1) c) Claim: The series ∞ n=1 3n is convergent. Proof: We apply the root test. In this case, an = P n2 3n , so 2 nn . |an | = 3 1 n 1 2 1 1 We have shown that (n n ) converges to 1, so the multiplication limit law implies that (n n ) = (n n · n n ) 2 converge to 1 · 1 = 1. Since (n n ) converges to 1, the scalar multiplication limit law then im1 2 1 plies that (|an | n ) = ( 31 n n ) converges to 13 (1) = 31 . Since lim |an | n = 13 , Theorem 10.7 implies P∞ n2 P 1 1 that α = lim sup |an | n = lim |an | n = 13 < 1. So the root test implies that ∞ n=1 3n n=1 an = P∞ n2 P a = converges absolutely, so by Corollary 14.7, ∞ converges. n=1 3n n=1 n Remark: One could also apply the ratio test here. In this case, we notice that (n + 1)2 |an+1 | = |an | 3n2 | = for all n. We may then use the limit laws to show that lim |a|an+1 n| |an+1 | |an | |an+1 | |an | 1 3 1 3 and Theorem 10.7 to show = lim = < 1. The Ratio test then implies that that lim sup is absolutely convergent, hence convergent. cos2 (n) is convergent. n2 |cos2 (n)| Proof: Notice that ≤ n12 for all n n2 2 P cos (n) test implies that is convergent. n2 P n 14.2) b) (−1) does not converge. e) Claim: The series P∞ n=1 an = n2 n=1 3n P∞ P ∈ N. Since P 1 n2 is convergent, the comparison Proof: Suppose that (−1)n converges. Then, by Corollary 14.5, (−1)n converges to 0. So, since .5 > 0, there exists N such that if n > N , then |(−1)n − 0| < .5. However, |(−1)n − 0| = 1 for all n, so |(−1)n − 0| > .5 for all n, which is a contradiction. Therefore, ((−1)n ) does not P converge to 0, and −1)n does not converge. Remark: It would also be fair to note that we previously proved in class that ((−1)n ) does not converge, hence does not converge to 0. P 2 1 14.2) f) The series nn is convergent. 1 1 Proof: If an = n1n , then |an | n = n1 for all n. Since lim n1 = 0, as we proved in class, (|an | n ) 1 1 converges to 0. Theorem 10.7 then implies that α = lim sup |an | n = lim |an | n = 0. Therefore, the P P 1 P P 1 root test implies that an = an = nn is absolutely convergent, so by Corollary 14.7, nn converges. Remark: We could also use the comparison test, by noting that n1n ≤ n12 for all n ∈ N. P P P P 14.5) a) Claim: If an = A and bn = B, then an + bn = A + B. P Proof: Let sn = a1 + · · · + an . Then, since an = A, lim sn = A. Let tn = b1 + · · · + bn . P Then, since bn = B, lim tn = B. If un = (a1 + b1 ) + · · · + (an + bn ), then un = sn + tn for all n ∈ N. Then, since (sn ) and (tn ) converge, the addition limit law implies that lim un = P lim sn + lim tn = A + B. But (un ) is the sequence of partial sums for the series an + bn , so P an + bn = A + B. b) Claim: If an = A and k ∈ R, then kan = kA. P Proof: Let sn = a1 + · · · + an . Then, since an = A, lim sn = A. Let P P tn = ka1 + ka2 + · · · kan = k(a1 + · · · + an ) = ksn P P be the sequence of partial sums for kan . Since an = A, we see that lim sn = A, so the scalar P multiplication limit law implies that (tn ) = (ksn ) converges to kA. Therefore, kan = kA. P P P c) Notice that if an = A and bn = B, it is not reasonable to conjecture that an bn = AB, P P P since the partial sum for an bn is not the product of the partial sums for an and bn . One may check that this equality does not hold in general by considering the case where an = bn = 21n . P Remark: In fact, an bn need not converge at all, which we may see by considering the n (−1) √ case an = bn = ( n ). One may use the alternating series test from section 15, which is not P covered on the exam, to show that an converges. 14.6)a) Claim: If |an | is convergent and (bn ) is bounded sequence, then an bn is convergent. Proof: Since (bn ) is bounded, there exists M so that |bn | ≤ M for all n ∈ N. Exercise 14.5 b) P P implies that, since |an | is convergent, M |an | is also convergent. Then, |an bn | ≤ M |an | for all P P n ∈ N, so since M |an | is convergent, the comparison test implies that an bn is convergent. P Remark: One may also verify that an bn satisfies the Cauchy criterion as suggested by the P hint. We also note that it does not suffice to assume that an converges, as we can see by n (−1) examining the case an = bn = ( √n ) as in the remark after the previous exercise. P P