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B) Wir definieren nun induktiv eine Teilfolge (ank )k∈N mit ank ∈ Ik f¨ur alle k ∈ N. Induktionsanfang. h. an0 = a0 . § 5 Das Vollst¨andigkeits-Axiom 50 Induktionsschritt k → k + 1. Da in dem Intervall Ik+1 unendlich viele Glieder der Folge (an ) liegen, gibt es ein nk+1 > nk mit ank+1 ∈ Ik+1 . c) Wir beweisen nun, dass die Teilfolge (ank ) konvergiert, indem wir zeigen, dass sie eine Cauchy-Folge ist. Sei ε > 0 vorgegeben und N so groß gew¨ahlt, dass diam(IN ) < ε. Dann gilt f¨ur alle k, j N ank ∈ Ik ⊂ IN und an j ∈ I j ⊂ IN .

Es gilt 0 xb−k < b, also gibt es eine ganze Zahl a−k ∈ {0, 1, . , b − 1} und eine reelle Zahl δ mit 0 δ < 1, so dass xb−k = a−k + δ. Mit ξ−k := δbk erh¨alt man x = a−k bk + ξ−k mit 0 ξ−k < bk . Das ist die Behauptung f¨ur n = −k. Induktionsschritt n → n + 1. Es gilt 0 ξn bn+1 < b, also gibt es eine ganze Zahl an+1 ∈ {0, 1, . , b − 1} und eine reelle Zahl δ mit 0 δ < 1, so dass ξn bn+1 = an+1 + δ. d. Bemerkung. Die S¨atze 4 und 5 sagen insbesondere, dass sich jede reelle Zahl durch einen (unendlichen) Dezimalbruch darstellen l¨asst und umgekehrt.

F¨ur die Partialsummen gilt nach §1, Satz 6 sn = n ∑ xk = k=0 1 − xn+1 . d. 10) Beispiele. F¨ur x = ± 12 erh¨alt man die beiden Formeln 1 1 1 1 1 = 2, 1+ + + + +... = 2 4 8 16 1 − 1/2 1 1 1 1 1 2 1− + − + ∓... = = . 2 4 8 16 1 + 1/2 3 § 4 Folgen, Grenzwerte 39 Satz 7 (Linearkombination konvergenter Reihen). Seien ∞ ∑ an n=0 ∞ und ∑ bn n=0 zwei konvergente Reihen reeller Zahlen und λ, µ ∈ R. Dann konvergiert auch die Reihe ∑∞ n=0 (λan + µbn ) und es gilt ∞ ∞ ∞ n=0 n=0 n=0 ∑ (λan + µbn) = λ ∑ an + µ ∑ bn .

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