18.014 Calculus with Theory, Problem Set 11 Solutions

18.014 Problem Set 11 Solutions Total: 32 points Problem 1: Prove that a sequence converges if, and only if its liminf equals its limsup. Solution (4 points) Suppose {an } is a sequence that converges to a limit L. Then given � > 0, there exists an integer N such that n > N implies |an − L| < 2� . In particular, the set {an } for n > N is bounded below by L − 2� and bounded above by L + 2� . Thus, L+ � � � � ≥ inf an ≥ L − and L + ≥ sup an ≥ L − . 2 n≥N +1 2 n≥N +1 2 2 In particular, given � > 0, there exists N such that m > N implies � � � � � � � � � inf an − L� < �, � sup an − L� < �. � � � � n≥m n≥m We conclude lim inf an = L = lim sup an and the liminf equals the limsup. Conversely, suppose {an } is a sequence such that lim inf an = L = lim sup an . We will show limn→∞ an = L. Given � > 0, there exist N1 and N2 such that m > N1 implies � � � � � inf an − L� < � �n≥m � and m > N2 implies � � � � � sup an − L� < �. �n≥m � Let N = max{N1 , N2 }. If m > N1 , N2 , then L − � < inf an ≤ am ≤ sup an < L + �. n≥m n≥m In particular, |am − L| < � if m > N . Thus, limn→∞ an = L. 1 Problem 2: Use this fact to prove every Cauchy sequence of real numbers converges. Solution (4 points) First, recall the following lemma: Lemma: Every decreasing sequence that is bounded below converges, and every increasing sequence that is bounded above converges. Now, let {an } be a Cauchy sequence of real numbers. Then there exists M such that n, m > M implies |an − am | < 1. Putting m = M + 1, we observe aM +1 − 1 ≤ an ≤ aM +1 + 1 if n > M . Put C = max{aM +1 , a1 , . . . , aM } and put B = min{aM +1 − 1, a1 , . . . , aM }. Then B ≤ an ≤ C for all n. In particular, if bn = supm≥n am and cn = inf m≥n am , then {bn } is a decreasing sequence bounded below by B and {cn } is an increasing sequence bounded above by C. Thus, by the lemma {bn } converges to L1 and {cn } converges to L2 . To finish the proof, we again use that {an } is a Cauchy sequence. Given � > 0, there must exist N such that n, m > N implies |an − am | < 5� . Moreover, there must exist M1 > N such that |bM1 − L1 | < 5� and M2 > N such that |cM2 − L2 | < 5� . We can do this because limn→∞ bn = L1 and limn→∞ cn = L2 . Choose n ≥ M1 such that |an − bM1 | < 5� and choose m ≥ M2 such that |am − cM2 | < 5� . We can do this because bM1 = supk≥M1 ak and cM2 = inf k≥M2 ak . Now, we observe |L1 − L2 | ≤ |L1 − bM1 | + |bM1 − an | + |an − am | + |am − cM2 | + |cM2 − L2 | < �. Since this is true for every � > 0, we conclude that L1 = L2 . But, by the previous problem, if the liminf and the limsup are equal, then limn→∞ an exists. Therefore, every Cauchy sequence converges. Problem 3: Suppose the series �∞ n=1 an converges. Then limn→∞ an = 0. � Solution (4 points) Define sm = m n=1 an . Then by definition {sm } converges. In particular, {sm } is a Cauchy sequence (Problem 5 on the last practice exam). Thus, given � > 0, there exists N such that m, n > N implies |sn − sm | < �. If we choose n = m + 1, then we get |am | = |sm+1 − sm | < � whenever m > N + 1. We conclude limm→∞ am = 0. 2 Problem 4: A function f on R is compactly supported if there exists a constant B > 0 such that f (x) = 0 if |x| ≥ B. If f and g are two differentiable, compactly supported functions on R, then we define � ∞ (f ∗ g)(x) = f (x − y)g(y)dy. −∞ Prove (i) f ∗ g = g ∗ f and (ii) f � ∗ g = f ∗ g � . Solution (4 points) a) Using the substitution u = x − y, we have � t � x−t f (x − y)g(y)dy = − � x+t f (u)g(x − u)du = −t f (u)g(x − u)du. x+t x−t Using that f is compactly supported, choose B such that f (u) = 0 if |u| > B. Thus, if t > B + |x|, then � x+t � −B f (u)g(x−u)du = x−t B � f (u)g(x−u)+ � f (u)g(x−u)du+ −B x−t x+t f (u)g(x−u)du. B The first and third terms are zero since f (u) is zero whenever u < −B or u > B. Hence, our integral becomes � B f (u)g(x − u)du. −B Similarly, t � � B g(x − u)f (u)du = −t g(x − u)f (u)du −B if t > B. And we have � x+t (f ∗ g)(x) = lim t→∞ f (u)g(x − u)du = f (u)g(x − u)du −B x−t � t g(x − u)f (u)du = (g ∗ f )(x). = lim t→∞ B � −t b) Integration by parts tells us � t � �t � � f (x − y)g(y)dy = −f (x − y)g(y)� + −t −t 3 t −t f (x − y)g � (y)dy. The limit of the first term on the right as t → ∞ is lim (−f (x − t)g(t) + f (x + t)g(−t)) = 0 t→∞ since g(t) = 0 and g(−t) = 0 if t > B � for some B � > 0. Thus, � t � t � � (f ∗ g)(x) = lim f (x − y)g(y)dy = lim f (x − y)g � (y)dy = (f ∗ g � )(x). t→∞ t→∞ −t −t Applying part (a), we get (f � ∗ g)(x) = (f ∗ g � )(x) = (g � ∗ f )(x) as desired. Problem 5: Determine whether the series diverge, converge conditionally, or con­ verge absolutely. � �n √ ∞ ∞ � � n 2n + 100 n n (a) (−1) (b) (−1) . n + 100 3n + 1 n=1 n=1 √ x Solution (4 points) (a) Consider the function f (x) = x+100 . Note √ 1 x � f (x) = √ . − 2 x(x + 100) (x + 100)2 One observes f � (x) < 0 if x > 100. Hence, f is monotonically decreasing when x > 100. Moroever, its easy to see limx→∞ f (x) = 0. Now, we break up our sum into √ √ 100 ∞ � � n n n n (−1) + (−1) . n + 100 n=101 n + 100 n=1 The first term is a finite sum and the second term converges by Leibniz’s rule (Thm. 10.14). Thus, our series converges. √ n However, our series does not converge absolutely. To see this, let an = n+100 , � 1 a n bn = √x , and note limn→∞ bn = 1. By example one on page 398, we know that bn � diverges. Hence, by theorem 10.9, an diverges as well. � 2n+100 �n � 2 �n (b) This sum converges absolutely. Let a = and b = . Observe n n 3n+1 3 � an limn→∞ bn converges since it is a geometric series. Hence, by theorem � bn = 1 and 10.9, an converges as well. � Problem 6: Prove ∞ n=1 an converges absolutely if an = 1/n if n is a square and an = 1/n2 otherwise. 4 � Solution (4 points) Let sN = N n=1 an be the nth partial sum. Note � � 1 � 2 1 ≤ sN = + . 2 n2 n2 √ m n≤N n≤N m≤ N not a square � � 1 But, n≤N n22 ≤ 2 ∞ n=1 n2 . This is a finite number, C, by example one on page 398. Since the partial sums sN are an increasing sequence, bounded by C, they must converge by our lemma in problem 2. � �∞ 2 Problem 7: (a) Prove that�if ∞ then n=1 |an | converges, n=1 an converges. Give �∞ 2 a counterexample in which ∞ a converges but |a | n diverges. n=1 n n=1 ∞ c � (n!) (b) Find all real c for which the series converges. (3n)! n=1 � Solution (4 points) (a) Suppose ∞ n=1 |an | converges. By problem 3, we must have limn→∞ |an | = 0. Thus, there exists N such that |an | < 1 whenever n > N . In particular, we see a2n = |an |2 < |an | if n > N . Splitting up our series, we have ∞ � n=1 a2n = N � a2n + n=1 ∞ � a2n . n=N +1 The first sum is finite � because it is a finite sum. We may compare the second series term by term to ∞ n=N +1 |an |, which converges by hypothesis. �∞ � 1 �2 On the other hand, converges by example one on the top of page 398. n=1 n � 1 Yet, ∞ is the divergent harmonic series. n=1 n (b) Let bn = (n!)c . (3n)! First, we apply the ratio test, and we get bn+1 (n + 1)c = lim . n→∞ (3n + 3)(3n + 2)(3n + 1) n→∞ bn lim This limit is zero and the series converges if c < 3. The limit is ∞ and the series diverges if c > 3. For c = 3, we analyze each term. Note n n � (n!)3 k � k 1 =1· ≤ n (3n)! n + k k=1 2n + k 2 k=1 � 1 since 2k ≤ n + k and k ≤ 2n + k. But, ∞ n=1 2n converges because it is a geometric series. Thus, by the comparison test (Thm. 10.8), we conclude that our series converges when c = 3. 5 � 1 Problem 8: (a) Prove that limn→∞ pn k=qn k = log(p/q). (b) Show the series 1 + 1/3 + 1/5 − 1/2 − 1/4 + 1/7 + 1/9 + 1/11 − 1/6 − 1/8... converges to log 2 + 12 log(3/2). Solution (4 points) (a) We let � > 0. First choose N such that n ≥ N . Then for all n ≥ N , 1 pn < �/2 for all �� pn pn−1 � 1 � 1 1 − = < �/2. k k=qn k pn k=qn Now, choose M such that p−q < �/2 for all n ≥ M . As the function f (x) = 1/x pqn is monotonically decreasing, we get the estimate � pn−1 � 1 − k k=qn � pn qn dx x � ≤ 1 1 p−q − = < �/2. qn pn npq � pn Now, choose Ñ = max N, M and observe qn dx = log(x)|pn qn = log(p/q). Thus, x for all n ≥ Ñ , the triangle inequality and our work above implies: � pn � � pn � � � 1 � �� pn−1 � 1 �� 1 � � � 1 pn−1 � � �+� − log(p/q)�� ≤ �� − log(p/q)�� < �/2 + �/2 = �. − � � � k k k=qn k k k=qn k=qn k=qn (b) We begin by observing that s5m = 3m � k=1 2m � 1 1 − . 2k − 1 k=1 2m Now, 3m � k=1 6m 3m 6m 3m 3m �1 � 1 �1 �1 � 1 1 = − = − + 2k − 1 k=1 k k=1 2k k 2k k k=1 k=1 k=1 and thus s5m � � 6m 3m 6m 3m � � 1 1 � 1 1 1 � 1 1 1 = + = + + + . k 2 k k 2 k 3m 2m k=3m+1 k=2m+1 k=3m k=2m Thus limm→∞ s5m = log(6m/3m) + 12 log(3m/2m) = log 2 + 12 log(3/2). 6