Übungsblatt 6. Analysis 1, HS14

Transkript

1 Übungsblatt 6 Analysis, HS4 Ausgabe Donnerstag, 6. Oktober. Abgabe Donnerstag, 23. Oktober. Bitte Lösungen bis spätestens 7 Uhr in den Briefkasten des jeweiligen Übungsleiters am J- oder K-Geschoss von Bau Y27 legen. Übung. Sei α (0, ) und (a n ) n 0 eine Folge komplexer Zahlen so, dass für alle n N gilt a n+2 a n+ α a n+ a n. (i) Zeigen Sie: n N : a n+ a n α n a a 0. (ii) Zeigen Sie für n N, m N, m < n gilt: a n a m n (iii) Folgern Sie aus ii.), dass (a n ) n 0 konvergiert. a k+ a k αm α a a 0. Lösung. (i) We use induction. The induction statement is A(n) : a n+ a n α n a a 0. The case A(0) is clear. Now assume that A(n) holds for some arbitrary, but fixed n N. a n+2 a n+ α a n+ a n α(α n a a 0 ) = α n+ a a 0 Where we first used the given property of (a n ) and then the assumption that A(n) is true. Hence, A(n+) is true. By the principle of induction we conclude that our claim holds.

2 (ii) Let n, m N with m n. If n = m it s clear. Let now m < n. We calculate a n a m = n a k+ a k n a k+ a k n α k a a 0 a a 0 α k k m Where we used Satz (ii) (n-m+ times) for the first inequality and Übung (i) for the second inequality. As α (0, ) we may apply Beispiel 6..2 and obtain a a 0 k m α k = a a 0 α m k α k = a a 0 αm α. (iii) First we treat the case when a a 0 = 0. In this case we get bei (ii.) n N : a n a 0 α a a 0 = 0 n N : a n = a 0. Hence, our sequence is constant and therefore convergent. Let now a a 0 0. Let ε > 0 arbitrary, but fixed. Since α (0, ) we know from Beispiel that α n converges. By Bemerkung we get α n 0. Hence, we can choose N N such that m N : α m (ii) we get: n m N : a n a m a a 0 As a n a m = a m a n we conclude: α ε a a 0 2 α ε a a 0 2 α = ε 2 < ε.. Using Übung n, m N : a n a m < ε. Hence, (a n ) n 0 is a Cauchy sequence and converges therefore by Satz Übung 2. Sei (a n ) eine monoton fallende Folge nichtnegativer reeller Zahlen. Wir definieren S n := n a i und T n := n 2 i a 2 i für n N. i= i= (i) Zeigen Sie für alle n N die folgenden Ungleichungen (verwenden Sie die - Notation): a) Sei n N und sei K N so, dass n {2 K, 2 K +,..., 2 K+ }. Dann gilt: b) Sei K N. Dann gilt: (ii) Folgern Sie aus Übung 2 i.), dass gilt: S n a + T K. T K 2S 2 K. a n konvergiert 2 n a 2 n konvergiert. 2

3 (iii) Entscheiden Sie (mit Beweis) für jedes α R, ob die Reihe n α konvergiert. Lösung 2. The convergence criterion from (ii.) is known as Cauchy condensation test (Cauchy Verdichtungskriterium). (i) a) S n = n a k = a + n a + K j= 2 j+ a k a + 2K+ j=2 a k = a + K a 2 j = a + K 2 j a 2 j = a + T K. j j= j= 2 j+ a k j For the second inequality we used that (a n ) is nonnegative and monotonically decreasing. b) T K = K 2 i a 2 i = 2 K i= 2 i i= j=2 i + a 2 i 2 K 2 i i= j=2 i + a j = 2 2K a i = 2S 2 K. Where we once again used that (a n ) is monotonically decreasing. i= (ii) By assumption holds n N \ {0} : a n 0. Therefore, both (S n ) and (T n ) are monotonically increasing. If a n converges, then by definition 6... (S n ) converges. By Lemma (S n ) is bounded and hence by Übung 2 (i.) b.) (T n ) is bounded. With Satz we conclude that 2 n a 2 n converges. Now we prove the other implication. If 2 n a 2 n converges, then by definition 6... (T n ) converges. By Lemma (T n ) is bounded and hence by Übung 2 (ii.) b.) (S n ) is bounded. With Satz we conclude that a n converges. (iii) From Übung 2 (ii.) we know that n α converges if and only if 2 n (2 n ) α = 2 n( α) = (2 α ) n 2 n (2 n ) α does. From Beispiel we know that for q R hold: q n converges if and only if q [0, ). As 2 α is positive for all α R, we get that 2 n (2 n ) converges if α and only if 2 α <. With Übung 2 (ii.) we finally obtain: n α converges α (, ). 3

4 Übung 3. Seien (a n ), (b n ) komplexe Folgen. Wir definieren B n := n b i für n N. (i) Zeigen Sie für alle n N die folgende Identität: i= n a i b i = a n+ B n + n (a i a i+ )B i. i= i= (ii) Sei nun (a n ) eine monoton fallende reelle Nullfolge (siehe Bemerkung 6.2..) und (b n ) sei eine komplexe Folge so, dass (B n ) eine beschränkte Folge ist. Nutzen Sie Übung 3 i.) um zu zeigen, dass a n b n konvergiert. Hint: Beweisen Sie, dass a n+ B n 0 und folgern Sie daraus, dass a n+ B n 0. (iii) Folgern Sie aus Übung 3 ii.) das Leibnizkriterium (Satz ). Lösung 3. (i) = n a n+ B n + n (a k a k+ )B k = a n+ B n + n a k B k n a k+ B k a k B k n a k+ B k = n a k B k n a k B k = a B + n a k (B k B k ) = a b + n a k b k = n a k b k. (ii) Note that a monotonically decreasing sequence can only converge to zero if it consist of non-negative numbers. Hence, we have n N \ {0} : a n 0. By Satz and Übung 3 (i.) it s enough to show that (a n+ B n ) and ( n (a k a k+ )B k ) converge. For the first term we proceed as follow. By assumption (B n ) is bounded. Hence, Therefore, we have for all n N \ {0} M [0, ) n N : B n M 0 a n+ B n = a n+ B n M a n+ = Ma n+ 4

5 As a n 0 we get with Bemerkung that Ma n = M 0 = 0. With Satz and the above calculation we obtain that a n+ B n 0. We rewrite this in a formal way and obtain: a n+ B n 0 ε > 0 N > 0 n > N : a n+ B n 0 < ε ε > 0 N > 0 n > N : a n+ B n < ε a n+ B n 0. Hence, (a n+ B n ) converges (to 0). For the convergence of the second term it s by Bemerkung enough to show ( n ) that a k a k+ B k converges. As (a n) is nonnegative and monotonically decreasing and (B n ) is bounded we get: n a k a k+ B k = n (a k a k+ ) B k n (a k a k+ )M = M(a a n+ ) Ma. ( n ) Hence, a k a k+ B k is bounded and monotonically increasing and therefore by Satz convergent. (iii) Let (a n ) a monotonically decreasing sequence of real numbers. Define b n := { ( ) n n 0, n gerade. It s clear that b k =, n ungerade. So (B n ) is bounded. Hence, we may apply Übung 3 (ii.) and obtain that ( ) n a n converges. This is exactly the statement of the alternating series test (Leibniz-Kriterium) Satz Übung 4. Zeigen Sie, dass die folgenden Reihen konvergieren. (i.) (ii.) (iii.) 8n i n ln(n+), wobei i C mit i2 = (siehe Bemerkung ). Hint: Berechnen Sie (iv.) n i k für n {, 2, 3, 4, 5}. 2 n n!, wobei n! in Definition.3.. definiert ist. 5 n n n 5

6 Lösung 4. (i) As 8n n 2 and we know from Übung 2 (iii.) that n 2 converges we conclude with the direct comparison test (Majorantenkriterium) Korollar that 8n converges. (ii) Let l N and n {0,, 2, 3}. Note that i 4l = (i 4 ) l = l =. 4l+n i k = n i 4l+k + l (i 4k + i 4k+ + i 4k+2 + i 4k+3) = n i k + l ( + i + i 2 + i 3) = n i k + l ( ) + i i = n i k We obtain: 0, n = 0 mod 4 n i k i, n = mod 4 = i, n = 2 mod 4, n = 3 mod 4 ( n Hence, decreases to zero. So (iii) We define a n := 2n n!. a n+ a n = 2 n+ (n+)! 2 n n! i k) is bounded by the above calculation and ( ln(n+) ) monotonically i n ln(n+) = 2n+ 2 n n! (n+)! = 2 n+. converges by Übung 3 (iii.). From Beispiel we know n 0. With Bemerkung we get 2 n 2 0 = 0. Hence, we get a n+ a n 0. By the ratio test (Quotientenkriterium) Satz we conclude that 2 n n! converges. (iv) We define a n := 5n n. n n an = 5 n. By the argument from above, we get n a n 0. By Lemma we get lim inf n a n = lim n a n = 0. Hence, by the root test (Wurzelkriterium) Satz we conclude that 5 n n n converges. 6