Integral Test: Convergence/Divergence of $\sum_{x=2}^\infty (\ln x)^{-1}$

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Homework Help Overview

The discussion revolves around determining the convergence or divergence of the series \(\sum_{x=2}^{\infty} (\ln x)^{-1}\) using the integral test. Participants are exploring the properties of the integral associated with the series and its behavior as \(b\) approaches infinity.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the integrability of the function \((\ln x)^{-1}\) and whether the integral diverges or if finding an antiderivative is the issue. There are suggestions to use comparison tests with other functions, such as \(1/\sqrt{x}\) and \(1/n\), to analyze divergence.

Discussion Status

The discussion is ongoing, with participants raising questions about the nature of the integral and exploring different approaches to establish divergence. Some guidance has been offered regarding comparison tests, but there is no explicit consensus on the method to be used.

Contextual Notes

There is mention of the original poster's uncertainty regarding the integral's behavior and the potential for using alternative tests due to difficulties in finding an antiderivative. The participants are also considering their backgrounds in calculus, which may influence their approaches.

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Homework Statement

Determine convergence or divergence using the integral test for.
[tex]\sum _{x=2}^{\infty } \left( \ln \left( x \right) \right) ^{-1}[/tex]

Homework Equations

i should take the limit as b goes to infinity of

[tex]\int _{2}^{b}\! \left( \ln \left( x \right) \right) ^{-1}{dx}[/tex]

The problem is that the function is not integrable.I know there was trick show that it is divergent but I don't remember what it was.

The Attempt at a Solution

 
Last edited:
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When you say it isn't integrable, do you mean the integral diverges or that you can't find an antiderivative?

The integral and the series converge or diverge togther.
 
LCKurtz said:
When you say it isn't integrable, do you mean the integral diverges or that you can't find an antiderivative?

The integral and the series converge or diverge togther.

I can't find the antiderivative. Maple give's me [tex]-{\it Ei} \left( 1,-\ln \left( x \right) \right)[/tex] and i don't know what it means. I am currently doing cal 3 but I think remember this exercise from cal 2. Is there any other test that I could use to show the divergence?
 
I would do a comparison with 1/√x.
 
ln(2n) = n*ln(2) < n so 1/ln(2n) > 1/n if that helps.
 
Bohrok said:
I would do a comparison with 1/√x.

Thanks. It is divergent by the comparison test.
 

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