Why does 1/[nlog(n+1)] diverge

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Homework Help Overview

The discussion centers around the divergence of the series 1/[nlog(n+1)], with participants exploring its behavior in comparison to the known divergent series 1/[nlog(n)].

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants investigate the integral of a related function and question the applicability of the integral test to the specific series in question. There is a discussion about the differences between the series involving log(n) and log(n+1), with some suggesting a comparison test as a potential approach.

Discussion Status

The discussion is active, with participants raising questions about the validity of certain tests and exploring different interpretations of the series. Some guidance has been offered regarding the comparison test, but no consensus has been reached.

Contextual Notes

Participants note the specific form of the series, emphasizing the distinction between log(n) and log(n+1), which may affect the application of convergence tests.

grossgermany
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Homework Statement



Why does the series 1/[nlog(n+1)] diverge

Homework Equations


We know that 1/[nlog(n)] diverges by the integral test. However the question as written does not lend itself to be any integral precisely.


The Attempt at a Solution

 
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investigate the integral:

[tex] f(x) = \int_{1}^{x}{\frac{dt}{t \, \log{t}}}[/tex]

as [itex]x \rightarrow \infty[/itex].
 
Yes, however the integral doesn't work in this particular case
Note that the question says nlog(n+1), not nlogn
 
grossgermany said:
Yes, however the integral doesn't work in this particular case
Note that the question says nlog(n+1), not nlogn

1/((n+1)*log(n+1)) diverges, right? You can do an integral test or just notice it's the same series as 1/(n*log(n)) shifted one term. Now try and think of how to use a comparison test.
 

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