Math_Analysis_HW4_Solutions.pdf - Real Analysis Homework IV...

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Unformatted text preview: Real Analysis Homework IV solutions October 11th, 2019 Exercise 2.3.3 Suppose {xn } is a bounded sequence and {xnk } is a subsequence. Prove lim inf xn ≤ lim inf xnk . n→∞ k→∞ Proof. Let an = inf{xm : m ≥ n}, bn = inf{xnm : m ≥ n}. Then, lim an = lim inf xn , lim bn = lim inf xnk For any n, consider the sets Sn = {xm : m ≥ n} and Tn = {xnm : m ≥ n}. Notice that nm ≥ m (this was proven in class), so if m ≥ n, then nm ≥ n . It follows that if xnm ∈ Tn , then xnm ∈ Sn , so Tn ⊂ Sn . By Exercise 1.1.4, this means inf Sn ≤ inf Tn , or an ≤ bn . Since n was arbitrary, we can take the limit (Prop 2.2.3) to conclude that lim inf xn ≤ lim inf xnk . n→∞ k→∞ Exercise 2.3.5 a. Let xn = (−1)n , n b. Let xn = (n−1)(−1)n , n (−1)n a. Since n ≤ find lim sup xn and lim inf xn . 1 n find lim sup xn and lim inf xn . and 1 n (−1)n → 0, Prop 2.2.10 implies n → 0. Hence, by Prop. 2.3.5, 0 = lim (−1)n (−1)n (−1)n = lim sup = lim inf . n n n 1 b. Let an = {sup{xk : k ≥ n}, then lim sup xn = lim an . We claim that for all n, an = 1 . To prove this, we first note that 1≥ (n − 1)(−1)n n so 1 is an upper bound . On the other hand, if b < 1 , then by the Archimedean property there exists an m such that 1 <1−b m Then, taking k = 2(m + n) , we see that k > n , so xk ∈ {xk : k ≥ n} , and furthermore, (k − 1)(−1)k k 1 =1− k 1 ≥1− m >b xk = so b is not an upper bound and an = 1. It follows that 1 = lim an = lim sup xn . Similarly, let bn = inf{xk : k ≥ n}. We claim bn = −1 for all n. To see this, observe that −1 is a lower bound of {xk : k ≥ n} for all n, and that for any b > −1 there is an m such that 1 < b − (−1), m and hence for k = 2(m + n) + 1, (k − 1)(−1)k k 1 = −1 + k <b xk = so −1 = inf{xk : k ≥ n} = bn . Thus, lim inf xn = lim bn = −1. 2 Exercise 2.4.1 Prove that n n2 −1 n2 o is a Cauchy sequence using directly the definition of Cauchy sequences. Proof. Let  > 0 be arbitrary. Then, since n22 → 0 , there exists an M such that for all n ≥ M , 2 <  . Choose M , and let n, k ≥ M . n2 Without loss of generality, assume n ≥ k. Then, n2 − 1 k 2 − 1 n2 − k 2 − = n2 k2 n2 k 2 (n − k)(n + k) ≤ n2 k 2 n(2n) ≤ 2 2 nk 2 ≤ 2 k < Exercise 2.5.3 Decide the convergence or divergence of the following series. P∞ 3 a. n=1 9n+1 P∞ 1 b. n=1 2n−1 c. P∞ d. P∞ e. P∞ n=1 (−1)n n2 1 n=1 n(n+1) n=1 2 ne−n We make the following observation: P P Lemma . If xn diverges and α 6= 0, then αxn diverges. Proof. We prove P the contrapositive. Suppose αxn converges. Then, by proposition 2.5.12(i), X xn = X1 1X αxn = αxn α α converges. 3 12 3 a. Observe that 4 9n+1 = 9n+1 ≥ n1 . Thus, by example 2.5.11 and the comparison test, 4 9n+1 P 3 diverges. By our lemma, this implies diverges as well. 9n+1 3 b. We have that diverges. 1 2n−1 ≥ 11 . 2n By the lemma and example 2.5.11, P11 2n diverges, so P 1 2n−1 P (−1)n P 1 P (−1)n c. Observe that converges by the p-test (Prop 2.5.17). Thus, n2 = n2 n2 converges absolutely, and so it converges (Prop. 2.5.15). P 1 d. Let SN = N n=1 n(n+1) . Then, SN = = N X n=1 N X n=1 =1− so lim SN = 1 , and P 1 n(n+1) 1 n(n + 1) 1 1 − n n+1 1 N +1 converges to 1 . e. We have that lim n→∞  ne−n 2 1/n = lim n1/n e−n n→∞    1/n −n = lim n lim e n→∞ n→∞ =1·0 =0 By Proposition 2.3.5, if limn→∞ xn exists, then lim supn→∞ xn = limn→∞ xn . Thus,  lim sup ne −n2 1/n =0<1 n→∞ so the series P ne−n converges by the root test (Prop 2.6.1). 2 Exercise 2.5.4 a. Prove that if P∞ n=1 xn converges, then P∞ n=1 (x2n + x2n+1 ) also converges. b. Give an explicit example where the converse does not hold. a. Proof. Let Sn = Tn = n X k=1 n X k=1 4 xk (x2k + x2k+1 ) P P∞ be the partial sums of ∞ x and We want to prove n n=1 n=1 (x2n + x2n+1 ), respectively. P that limn→∞ Tn exists. Observe limn→∞ Sn converges , since ∞ x n=1 n converges. Moreover, the partial sums are related by Tn = S2n+1 − x1 . Using the rules of limit arithmetic, lim Tn = lim S2n+1 − x1 = n→∞ so P∞ n=1 (x2n n→∞ ∞ X xn − x1 . n=1 + x2n+1 ) converges. b. Let xn = (−1)n . Then, ∞ ∞ X X (x2n + x2n+1 ) = 0=0 n=1 but n=1 ∞ X xn = n=1 ∞ X (−1)n n=1 cannot converge, since the terms do not go to zero (Prop 2.5.9). Exercise 2.5.6 Let P xn be a series a. If there is an N and a ρ < 1 such that |xn+1 | <ρ |xn | for all n ≥ N , then the series converges absolutely. b. If there is an N such that for all n ≥ N , |xn+1 | ≥1 |xn | then the series diverges. (Although it is not explicitly given, we will assume xn 6= 0 for n ≥ N in order to ensure all the fractions are well-defined). P a. Proof. By Prop 2.5.6, it is enough to prove that ∞ n=N +1 |xn | converges. For n ≥ N , |xn | |xn−1 | |xN +1 | ··· |xN | |xn−1 | |xn−2 | |xN | ≤ ρn−N |xN | P n−N By Prop P 2.5.5, 2.5.6 and Prop 2.5.12(i), ∞ |xN | converges, so by the comparin=N +1 ρ son test |xn | converges. |xn | = 5 b. Proof. Observe that for n ≥ N , |xn | |xn−1 | |xN +1 | ··· |xN | |xn−1 | |xn−2 | |xN | ≥ |xN | |xn | = Thus, |x Pn | ≥ |xN | > 0 for all n ≥ N , so xn does not converge to 0 . By Prop 2.5.9, this shows xn diverges. Exercise 2.5.8 Show that converges. P (−1)n The partial sums of ∞ are n=1 n P∞ n=1 (−1)n n Sn = n X (−1)k k=1 k . We wish to prove that {Sn } converges. First, let us consider S2n . By grouping terms, we find that n n X X 1 1 −1 S2n = − = 2k 2k − 1 k=1 2k(2k − 1) k=1 which are exactly the partial sums of ∞ X n=1 By comparison to the p-series 1 n=1 n2 , P∞ −1 . 2n(2n − 1) we see that this series converges, so lim S2n = n→∞ ∞ X k=1 −1 2k(2k − 1) exists. Let us call this limit S. Certainly, if Sn converges, it must converge to S. Let’s show lim Sn = S. Let  > 0 be arbitrary. Since S2m convergesto S, there exists an M such that for all m ≥ M , |S2m − S| < . 2 Choose M , and let N ≥ max  , 2M . Consider n ≥ N . There are two cases: either n is even or n is odd. If n is even, n = 2k. It follows that k ≥ M , so |S2k − S| <  2 by the choice of M . On the other hand, if n is odd, n = 2k − 1. Then, |Sn − S| ≤ |Sn − Sn+1 | + |Sn+1 − S| 6 As above, |S2k − S| < On the other hand,  2 (−1)n |Sn − Sn+1 | = n 1 = n 1 ≤ N  ≤ 2 so |Sn − S| ≤ |Sn − Sn+1 | + |Sn+1 − S| <  P (−1)n So, |Sn − S| <  and converges. n Exercise 2.5.9 a. Prove that if P xn and P yn converge absolutely, then P xn yn is absolutely convergent. b. Find an example where the converse is not true. c. Find P an explicit P example P where all three series are convergent, are not just finite sums, and ( xn ) ( yn ) 6= xn yn . That is, series are not multiplied term-by-term. P a. Since yn converges, yn → 0 by Prop 2.5.9, so yn P is bounded . Let B be such that |yn | ≤ B for all n . Then, |xn yn | ≤ B|xnP |. Since B|xn | converges (Prop 2.5.12(i) and |xn yn | converges as well, which shows that P the definition of absolute convergence), xn yn is absolutely convergent. b. Let ( 1 n is even xn = 0 n is odd ( 0 n is even yn = 1 n is odd P P P Then, xn yn = 0 is absolutely convergent. However, |xn | Pxn yn = 0 for all n, so and |yn | do not converge since |xn |, |yn | 6→ 0 (Prop 2.5.9). n c. Let xn = yn = 12 . Then, by Prop 2.5.5, X X X X |xn | = |yn | = xn = yn = 1 P P so xn , yn are absolutely convergent, but X X 1 1 X   X  xn y n = = 6= xn yn . 4n 3 7 Exercise 2.5.11 Prove that if an > 0 and bn > 0 for all n, and 0 ≤ lim then P an and P an ≤∞ bn bn either both converge or both diverge. Proof. By limit arithmetic, if then 0 ≤ lim an ≤∞ bn  −1 bn an 0 ≤ lim = lim ≤∞ an bn P so it is enough to prove that if bn converges, then an converges, since we can reverse the roles of an and to get the general result. P bn and take the contrapositive Suppose bn converges, and let L = lim abnn . Then, there exists an M such that for n ≥ M , an <L+1 bn P P∞ choose N . By Prop 2.5.6, ∞ n=N bn converges, so, since 0 < an < P(L + 1)bn , n=N an converges (Prop 2.5.12 and the comparison test). Finally, by Prop 2.5.6, an converges. Exercise 2.5.16 Use the Cauchy condensation principle to decide the convergence of P ln n a. n2 P 1 b. n ln n P 1 c. n(ln n)2 P 1 d. n ln n(ln ln n)2 P P P P To simplify notation, we will write an ∼ bn to mean an converges if and only if bn converges. Also, where necessary, we will star the summation at n = 2 or n = 3 instead of n = 1 to avoid undefined terms. a. When n ≥ 3, ln n increases slowly than n2 , so P lnmore P n ln n Prop 2.5.6, we see that ∼ ∞ n=3 n2 n2 ln n n2 is decreasing after n = 3. Using On the other hand, we can apply the Cauchy condensation test to the second series to obtain X ln n X n ln 2 ∼ n2 22n Since n ln 2 1/n 1 1 lim n = lim n1/n lim (ln 2)1/n lim = , n→∞ n→∞ n→∞ n→∞ 2 2 2 this last sequence converges, so the original sequence converges. 8 b. This sequence of terms is decreasing (here, we must begin with n = 2 to avoid undefined terms). Applying Cauchy condensation gives that X 1 1 ∼ n ln n n ln 2 P1 which diverges by comparison with , so the original series diverges. n X c. Since the terms in the series are decreasing (n ≥ 2), we can apply Cauchy condensation to get X X 1 1 ∼ 2 2 n(ln n) n (ln 2)2 P 1 which converges by comparison with the p-series , so the original series converges. n2 d. (We assume n ≥ 3 so all the terms are defined). The series is decreasing, so one application of condensation gives X X 1 1 ∼ n ln n(ln ln n n ln 2(ln(n ln 2))2 By logarithm arithmetic, ln(n ln 2) = ln n + ln ln 2. Thus, 1 1 1 1 = ≤ n ln 2(ln(n ln 2))2 n(ln n + ln ln 2)2 ln 2 n(ln n)2 so, by comparison with the series in part (c), the original sequence converges. 9 ...
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