MaPhy-&Atilde;œ7

# MaPhy-&Atilde;œ7 - Aufgabe 1 a Beh a k = 1 √ k → a...

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Unformatted text preview: Aufgabe 1 a) Beh.: a k = 1 √ k → a | a ∈ R Bew.: Sei ε > 0, nach Archimedes: ∃ k ∈ N | k > 1 √ k 1 √ k- < ε 1 ε < √ k 1 ε 2 < k dann gilt: ∀ k ≥ k : k ≥ k > 1 √ k ⇒ √ k > 1 ε ⇒ < 1 √ k < ε ⇒ 1 √ k- < ε ∀ k ≥ k ⇒ 1 √ k → b) Beh.: b k = √ k + 1- √ k → b | b ∈ R Bew.: ≤ b k = √ k + 1- √ k = √ k +1 2- √ k 2 √ k +1+ √ k = 1 √ k +1+ √ k ≤ 1 √ k → c) Beh.: c k = (- 1) k · k → c | c ∈ R Bew.: c k = (- 1) k · k ⇒ | c k | = k → ∞ ( n → ∞ ) ⇒ c k ist unbeschr¨ ankt da aber c 2 n +1 =- (2 k + 1) → -∞ c 2 k = +2 k → + ∞ ⇒ c k ist unbeschr¨ ankt divergent d) Beh.: d k = k √ k → d | d ∈ R Bew.: d k = k √ k = k 1 k = e 1 k log( k ) Es ist k ≤ (1+ ε ) k ε , dann(1 + ε ) k = n ∑ j =0 ( k j ) ε j ≥ kε ⇒ 1 ≤ k √ k ≤ k √ (1+ ε ) k k √ ε = (1 + ε ) 1 k √ ε |{z} → 1 → 1 + ε ⇒ 1 ≤ lim k →∞ k √ k ≤ lim k →∞ k √ k ≤ 1 + εn , ∀ ε > ⇒...
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## This note was uploaded on 01/18/2010 for the course FB07 BP-03 taught by Professor Lani-wayda during the Winter '09 term at Uni Giessen.

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MaPhy-&Atilde;œ7 - Aufgabe 1 a Beh a k = 1 √ k → a...

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