Anki Deck Changes

Commit: 6badd035 - add rho formula to cloze

Author: obrhubr <obrhubr+noreply@noreply.com>

Date: 2026-03-31T08:49:26+02:00

Changes: 1 note(s) changed (0 added, 1 modified, 0 deleted)

Note 1: ETH::2. Semester::Analysis

Deck: ETH::2. Semester::Analysis
Note Type: Horvath Cloze
GUID: ==Fn-80|4a
modified

Before

Front

ETH::2._Semester::Analysis::3._Reihen::3._Konvergenzkriterien
Wurzelkriterium: Sei \(\rho = \limsup_{n\to\infty} |a_n|^{1/n}\). Dann:
  • \(\rho < 1\) \(\implies\) \(\sum a_n\) konvergiert absolut
  • \(\rho > 1\) \(\implies\) \(\sum a_n\) divergiert
  • \(\rho = 1\) \(\implies\) keine Aussage möglich

Proof Included

Back

ETH::2._Semester::Analysis::3._Reihen::3._Konvergenzkriterien
Wurzelkriterium: Sei \(\rho = \limsup_{n\to\infty} |a_n|^{1/n}\). Dann:
  • \(\rho < 1\) \(\implies\) \(\sum a_n\) konvergiert absolut
  • \(\rho > 1\) \(\implies\) \(\sum a_n\) divergiert
  • \(\rho = 1\) \(\implies\) keine Aussage möglich

Proof Included
Wenn Quotientenkriterium versagt (\(\rho=1\)), versagt auch das Wurzelkriterium — aber nicht umgekehrt.

Proof: 
  1. Convergence \(L < 1\)
    1. \(\sum a_n \geq 0\), \(\displaystyle L = \limsup_{n\to\infty} \left| {a_n}^{1/n} \right| < 1\).
    2. Choose \(q\) with \(L < q < 1\). Since \(\limsup \left| {a_n}^{1/n} \right| = L\), there exists \(N\) such that for all \(n \geq N: \left| {a_n}^{1/n} \right| \leq\) \(q \implies \left| a_n \right| \leq q^n\)
    3. So \(\sum_{n=N}^{\infty} \left| a_n \right| \leq\) \(\sum_{n=N}^{\infty} q^n < \infty\) (geometric series, \(q < 1\)).
    4. Hence \(\sum \left| a_n \right|\) converges. (Majorantenkriterium)
  3. Divergence \(L > 1\)
    1. \(\sum a_n \geq 0\), \(\displaystyle L = \limsup_{n\to\infty} \left| {a_n}^{1/n} \right|\) \(> 1\).
    2. Since \(\limsup \left| {a_n}^{1/n} \right| > 1\), there exist infinitely many \(n\) such that: \(\left| {a_n}^{1/n} \right| >\) \(1 \implies |a_n| > 1\)
    3. So \(|a_n| \not\to 0\), hence \(\sum a_n\) diverges.Convergence 

After

Front

ETH::2._Semester::Analysis::3._Reihen::3._Konvergenzkriterien
Wurzelkriterium: Sei \(\rho = {{c4:: \limsup_{n\to\infty} |a_n|^{1/n} }}\). Dann:
  • \(\rho < 1\) \(\implies\) \(\sum a_n\) konvergiert absolut
  • \(\rho > 1\) \(\implies\) \(\sum a_n\) divergiert
  • \(\rho = 1\) \(\implies\) keine Aussage möglich

Proof Included

Back

ETH::2._Semester::Analysis::3._Reihen::3._Konvergenzkriterien
Wurzelkriterium: Sei \(\rho = {{c4:: \limsup_{n\to\infty} |a_n|^{1/n} }}\). Dann:
  • \(\rho < 1\) \(\implies\) \(\sum a_n\) konvergiert absolut
  • \(\rho > 1\) \(\implies\) \(\sum a_n\) divergiert
  • \(\rho = 1\) \(\implies\) keine Aussage möglich

Proof Included
Wenn Quotientenkriterium versagt (\(\rho=1\)), versagt auch das Wurzelkriterium — aber nicht umgekehrt.

Proof: 
  1. Convergence \(L < 1\)
    1. \(\sum a_n \geq 0\), \(\displaystyle L = \limsup_{n\to\infty} \left| {a_n}^{1/n} \right| < 1\).
    2. Choose \(q\) with \(L < q < 1\). Since \(\limsup \left| {a_n}^{1/n} \right| = L\), there exists \(N\) such that for all \(n \geq N: \left| {a_n}^{1/n} \right| \leq\) \(q \implies \left| a_n \right| \leq q^n\)
    3. So \(\sum_{n=N}^{\infty} \left| a_n \right| \leq\) \(\sum_{n=N}^{\infty} q^n < \infty\) (geometric series, \(q < 1\)).
    4. Hence \(\sum \left| a_n \right|\) converges. (Majorantenkriterium)
  3. Divergence \(L > 1\)
    1. \(\sum a_n \geq 0\), \(\displaystyle L = \limsup_{n\to\infty} \left| {a_n}^{1/n} \right|\) \(> 1\).
    2. Since \(\limsup \left| {a_n}^{1/n} \right| > 1\), there exist infinitely many \(n\) such that: \(\left| {a_n}^{1/n} \right| >\) \(1 \implies |a_n| > 1\)
    3. So \(|a_n| \not\to 0\), hence \(\sum a_n\) diverges.Convergence 
Field-by-field Comparison
Field Before After
Text Wurzelkriterium: Sei \(\rho = \limsup_{n\to\infty} |a_n|^{1/n}\). Dann:<br><ul><li>\(\rho {{c1::&lt; 1}}\) \(\implies\) \(\sum a_n\) konvergiert <b>absolut</b></li><li>\(\rho {{c2::&gt; 1}}\) \(\implies\) \(\sum a_n\) divergiert</li><li>\(\rho {{c3::= 1}}\) \(\implies\) <b>keine Aussage</b> möglich<br><br></li></ul><i>Proof Included</i> Wurzelkriterium: Sei \(\rho = {{c4:: \limsup_{n\to\infty} |a_n|^{1/n} }}\). Dann:<br><ul><li>\(\rho {{c1::&lt; 1}}\) \(\implies\) \(\sum a_n\) konvergiert <b>absolut</b></li><li>\(\rho {{c2::&gt; 1}}\) \(\implies\) \(\sum a_n\) divergiert</li><li>\(\rho {{c3::= 1}}\) \(\implies\) <b>keine Aussage</b> möglich<br><br></li></ul><i>Proof Included</i>
Tags: ETH::2._Semester::Analysis::3._Reihen::3._Konvergenzkriterien
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