Convergent and Divergent Sequences: The Relationship Between a_{n} and 1/a_{n}

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Homework Help Overview

The discussion revolves around the relationship between convergent and divergent sequences, specifically examining the implications of a sequence \( a_{n} \) converging or diverging, and the behavior of its reciprocal \( \frac{1}{a_{n}} \). Participants explore the conditions under which these sequences behave in relation to one another.

Discussion Character

  • Conceptual clarification, Assumption checking, Exploratory

Approaches and Questions Raised

  • Participants question the implications of a convergent sequence and its reciprocal, particularly whether divergence of \( \frac{1}{a_{n}} \) necessitates convergence of \( a_{n} \). There is also discussion about the terminology used, with some participants clarifying the distinction between sequences and series.

Discussion Status

The discussion is ongoing, with participants providing insights and suggestions for further exploration. Some guidance has been offered regarding the comparison of convergence between different series, though no consensus has been reached on the original question.

Contextual Notes

There is a mention of potential confusion due to terminology, specifically the use of "series" instead of "sequences." Additionally, the nature of the homework context is questioned, which may influence the depth of exploration in the discussion.

courtrigrad
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If we are given a convergent sequence [tex]a_{n}[/tex], then [tex]\frac{1}{a_{n}}[/tex] diverges. But what about the converse:
If [tex]\frac{1}{a_{n}}[/tex] diverges, then [tex]a_{n}[/tex] converges? [tex]a_{n} \rightarrow 0[/tex], so its possible that [tex]a_{n}[/tex] could converge. But it could also diverge, right?
 
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Take a_n=1.
 
ok thanks. so then it could both converge and diverge then.
 
You say "series" in the topic's title but you talk about sequences. And you say something that doesn't make sense to me: in my world, if a_n goes to a[itex]\neq[/itex]0, then 1/a_n goes to 1/a.

On the other hand if [itex]\sum a_n[/itex] converges, then [itex]\sum \frac{1}{a_n}[/itex] diverges. Is that what you meant?
 
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yes sorry for the bad terminology
 
Ok, then in this case, if the sequence [itex]1/a_n[/itex] diverges, then [itex]a_n \rightarrow 0[/itex] and although it is certain that [itex]\sum\frac{1}{a_n}[/itex] diverges, [itex]\sum a_n[/itex] may or may not converges.
 
Is this homework? If not, and you're just messing around to get a better handle on series, then I shouldn't have been so short (although you should probably post questions like this in the general math section). If you want to keep pursuing this line, try comparing the convergence of the sum of {a_n} with that of {(a_n)^p} for values of p other than -1, or even {f(a_n)} for arbitrary functions f. Can you find any conditions on p or f that allow you to conclude the convergence of the latter series from the former, or vice versa. Just a suggestion.
 

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