Convergence of Alternating Series: ln(1+x) <= x Hint for Absolute Convergence

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Homework Help Overview

The discussion revolves around the convergence properties of the series related to the function ln(1+x) for x > 0. Participants are exploring whether the series converges absolutely, conditionally, or diverges, with a specific hint suggesting the use of the inequality ln(1+x) <= x.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the nature of the series, noting that it appears to be alternating and questioning the implications of the hint regarding the inequality. There is also mention of the ratio test and the Leibniz test as potential methods for establishing convergence.

Discussion Status

The discussion is active, with participants providing insights into the convergence behavior of the series. Some suggest that the series is absolutely convergent, while others raise questions about the validity of this claim and the conditions necessary for convergence, particularly in the context of alternating series.

Contextual Notes

There is a hint provided that suggests establishing the inequality ln(1+x) <= x, which may be crucial for the analysis. Participants are also considering the implications of the terms of the series getting smaller and converging to zero, as well as the conditions required for applying the Leibniz test.

rcmango
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Homework Statement



Does this problem converge absolutely, conditionally, or does it diverge?

the equation: [/URL]

Homework Equations



also, the hint is to first show that ln(1 + x) <= x if x > 0

The Attempt at a Solution



It looks like an alternating series. not sure what the hint is implying or if its converging.

Thanks for any help.
 
Last edited by a moderator:
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Each term gets smaller and smaller, and converges to zero. It is absolutely convergent.

The ratio test will tell you it converges as well.
 
Last edited:
Gibz said:
Each term gets smaller and smaller, and converges to zero. It is absolutely convergent.
Is that what you mean to say? Each term of
[tex]\Sigma_{n\rightarrow \infty}\frac{1}{n}[/itex]<br /> "gets smaller and smaller, and converges to zero" but the series doesn't converge at all.[/tex]
 
when it is an alternating series you can use Leibniz test
your pic is not very clear... but my guess is that the hint is to help you establish one of the condition in the Leibniz test namely, the terms are getting smaller

Leibniz test:
If [tex]\sum_1^{\infty} (-1)^{n+1} b_n[/tex] such that all [tex]b_n>0[/tex] (ie alternating series) and [tex]b_{n+1} < b_n\; \forall\,n[/tex] and [tex]b_n\rightarrow 0[/tex], then series converges to S and [tex]|S-S_k|\leq b_{k+1}[/tex]
 

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