Proving Series Divergence: Convergence of a_n w/o Explicit Formula

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Homework Help Overview

The discussion revolves around proving the divergence of the series \(\sum \frac{a_{n}}{1+a_{n}}\) given that \(\sum a_{n}\) diverges, with the original poster expressing uncertainty about how to approach the problem without an explicit formula for the series.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the implications of the divergence of a series and consider the behavior of the terms \(a_{n}\) as \(n\) approaches infinity. There is discussion about bounding the terms of the series to analyze divergence.

Discussion Status

The conversation is ongoing, with participants sharing insights and questioning the relationship between the divergence of the original series and the behavior of its terms. Some guidance has been offered regarding bounding the terms, but no consensus has been reached yet.

Contextual Notes

There is a noted lack of explicit forms for the series, which participants acknowledge as a constraint in applying standard convergence tests.

CrusaderSean
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Given [tex]a_{n} > 0[/tex] and [tex]\sum a_{n}[/tex] diverges, show that [tex]\sum \frac{a_{n}}{1+a_{n}}[/tex] diverges.
Since I don't have an explicit form for the series, I can't apply any of the standard tests. I'm not sure where to start on this problem. I know the criteria for convergence/divergence, namely tail end of series has to converge or cauchy criterion condition. But I don't see how that helps without knowing what series looks like. Please steer me in the right direction.
 
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If a series diverges, what happens to to its reciprocal ?
 
I would say reciprocal converges, but apparently it's not enough to bring original series to convergence... I thought about this a little more and I think I'll analyze it based on how [tex]a_{n}[/tex] diverges. that is, does it go to zero, constant, or infinity as n goes to infinity and try to bound the reciprocal from below to show series diverges.
 
For [itex]a_n>1[/itex]:

[tex]\frac{a_n}{1+a_n}>\frac{1}{2}[/tex]

For [itex]a_n\leq 1[/itex]:

[tex]\frac{a_n}{1+a_n}\geq\frac{a_n}{2}[/tex]
 

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