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Cauchy-Kriterium für Folgen

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Exercise

https://texercises.com/exercise/cauchy-kriterium-fur-folgen/

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Exercise:
Eine reelle Folge ist genau dann konvergent wenn sie ein Cauchy-Folge ist.

Solution:
Beweis. Angenommen a_n_n ist eine reelle Folge mit Grenzwert A. Sei epsilon . Dann existiert ein N in mathbbN s.d. für alle n geq N gilt |a_n-A| fracepsilon. Für mn geq N gilt somit auch |a_m-a_n| leq |a_m-A|+|A-a_n| fracepsilon+fracepsilon epsilon Dies beweist dass a_n_n eine Cauchy-Folge ist. Sei umgekehrt a_n_n eine Cauchy-Folge. Für epsilon existiert dann ein N in mathbbN s.d. |a_m-a_n| für mn geq N. Insbesondere gilt also |a_n| leq |a_n-a_N|+|a_N| +|a_N| für alle n geq N. Daher ist a_n_n eine beschränkte Folge Lemma .. Des Weiteren existiert nach Annahme für jedes epsilon ein N in mathbbN s.d. |a_m-a_n| epsilon für mn geq N. Man setzt mN und erhält a_m-epsilon a_n a_m+epsilon. Man betrachte nun Liminf und Limsup der Folge und erhält a_m-epsilon lim textinf_n rightarrow infty a_n leq lim textsup_n rightarrow infty a_n a_m+epsilon. Insbesondere gilt |lim textinf_n rightarrow infty a_n-lim textsup_n rightarrow infty a_n| leq epsilon. Da aber epsilon beliebig erhält man Gleichheit von Limsup und Liminf und daher Konvergenz der Folge nach Korollar ..

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Tags	analysis i, beweis, cauchy, eth, folge, hs22, konvergent, proof
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Difficulty	(3, default)
Points	0 (default)
Language	(Deutsch)
Type	Proof
Creator	rk
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