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Why does the epsilon-N definition require the condition for every ε>0ε > 0?

The definition requires the condition for every ε>0\varepsilon > 0 to ensure that the sequence terms can be made arbitrarily close to the limit. A single band or a finite plotted sample is insufficient; the cutoff NN may depend on ε\varepsilon and need not be unique. All later terms, rather than just some selected terms, must satisfy the error bound.

Conditions

  • The tolerance ε\varepsilon is arbitrary and positive.
  • The cutoff NN depends on ε\varepsilon.
  • The condition must hold for all n>Nn > N.

Reasoning, step by step

  1. Recognize that a single tolerance band does not establish the full definition.
  2. Understand that the definition requires the condition for every ε>0\varepsilon > 0.
  3. Note that NN may depend on ε\varepsilon and need not be unique.
  4. Ensure that all later terms, not just selected ones, satisfy the error bound.
  5. Conclude that this universal quantification over ε\varepsilon guarantees convergence.

Example

The script states: 'The definition requires this for every ε>0ε>0, not just one band or a finite plotted sample. N may depend on ε and need not be unique. All later terms, rather than just some selected terms, must satisfy the error bound.'

Common misconceptions

  • Believing that one valid tolerance band proves convergence.
  • Thinking that NN must be the same for all ε\varepsilon.
  • Assuming that only some selected terms need to satisfy the error bound.

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