Can we Prove \lim_{n→∞}S_{n-1} = L Given \lim_{n→∞}S_{n} = L?

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Discussion Overview

The discussion revolves around the question of whether the limit of a sequence \( S_{n-1} \) approaches \( L \) given that the limit of the sequence \( S_n \) approaches \( L \). The scope includes mathematical reasoning and the application of limit definitions, particularly in the context of sequences and convergence.

Discussion Character

  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant questions the validity of deducing \( \lim_{n\to\infty} S_{n-1} = L \) from \( \lim_{n\to\infty} S_n = L \), noting that it is not immediately obvious.
  • Another participant suggests a substitution of variables, letting \( m = n-1 \), and argues that this should hold due to the nature of limits at infinity.
  • A later reply emphasizes the need for a more rigorous approach, specifically referencing the Cauchy limit condition.
  • One participant provides a detailed ε-N definition of limits to argue that if \( \lim_{n\to\infty} S_n = L \), then it follows that \( \lim_{n\to\infty} S_{n-1} = L \) based on the inequalities derived from the definition.

Areas of Agreement / Disagreement

Participants express varying degrees of certainty regarding the proof. While some suggest that the limit relationship holds through variable substitution, others seek a more rigorous justification. The discussion does not reach a consensus on the proof's validity.

Contextual Notes

Participants reference the Cauchy condition and ε-N definitions, indicating that the discussion may depend on these mathematical frameworks. There is an implicit assumption that the limits are defined in a standard manner, but this is not explicitly stated or agreed upon.

Bipolarity
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This might sound like a dumb question, but it's actually not too obvious to me. If we know that [itex]\lim_{n→∞}S_{n} = L[/itex], can we prove that [itex]\lim_{n→∞}S_{n-1} = L[/itex] ? I'm actually using this as a lemma in one of my other proofs (the proof that the nth term of a convergent sum approaches 0), but can't get around the proof of this not-so-obvious-but-still-quite-intuitive lemma.

I wrote down the Cauchy-definitions of both these limits, but have no idea how to deduce one from the other.

Thanks for all the help!

BiP
 
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Hey Bipolarity.

This seems like a bit of symbol shuffling but let m = n-1 and then consider lim m->infinity instead of lim n->infinity. Maybe I'm missing something though with what you are asking, but the nature of infinity should make the above hold.
 
chiro said:
Hey Bipolarity.

This seems like a bit of symbol shuffling but let m = n-1 and then consider lim m->infinity instead of lim n->infinity. Maybe I'm missing something though with what you are asking, but the nature of infinity should make the above hold.

Is there any way to be more rigorous, i.e. using the Cauchy limit?

BiP
 
That the limit exists implies that the Cauchy condition holds.
 
If you want a rigorous proof, using ε--N definition of limit, here it is.

If [itex]\lim_{n\to\infty} S_n =L[/itex], it means that for any [itex]\varepsilon>0[/itex] there exists [itex]N=N(\varepsilon)<\infty[/itex] such that for all [itex]n>N[/itex] the inequality [itex]|S_n-L|<\varepsilon[/itex] holds.

So, for any [itex]\varepsilon>0[/itex] the inequality [itex]n>N(\varepsilon)+1[/itex] implies that [itex]n-1>N(\varepsilon)[/itex], so [itex]|S_{n-1}-L|<\varepsilon[/itex], which means that [itex]\lim_{n\to\infty} S_{n-1} =L[/itex].
 

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