Determining convergence of (-1)^n/ln(n) as n goes to infinity

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Homework Help Overview

The discussion revolves around determining the convergence properties of the series (-1)^n/ln(n!). Participants are exploring whether it is divergent, conditionally convergent, or absolutely convergent, with references to various convergence tests.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Some participants mention the Alternating Series test and express uncertainty about proving absolute convergence. Others suggest using comparison tests, but there is confusion about appropriate comparisons. Questions arise regarding the reliability of Wolfram Alpha's outputs and the implications of its results.

Discussion Status

The discussion is ongoing, with participants sharing their thoughts on convergence tests and the challenges they face in proving their claims. Some have provided guidance on potential comparisons, while others express skepticism about the information from Wolfram Alpha.

Contextual Notes

There is mention of misleading outputs from Wolfram Alpha, which has contributed to confusion regarding the convergence of the series. Participants are also grappling with the implications of the integral test and the behavior of related functions.

FuzzieLogic
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Homework Statement



Determine whether (-1)^n/ln(n!) is divergent, conditionally convergent, or absolutely convergent.

Homework Equations


None, really? :S

The Attempt at a Solution



Okay, so I know the series converges by the Alternating Series test - terms are positive, decreasing, going to zero. I also know that it absolutely converges...because Wolfram Alpha told me so. =P I have no idea how to prove it though. The ratio test results in 1 so that's inconclusive, so I'm left with comparison or limit comparison test..but I don't know what to compare it to.

Thanks in advance for your help!
 
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I would compare it to 1/ln(n^n). That's less than 1/ln(n!), right? And 1/ln(n^n) diverges by an integral test. Do you have that?
 
FuzzieLogic said:
Okay, so I know the series converges by the Alternating Series test - terms are positive, decreasing, going to zero. I also know that it absolutely converges...because Wolfram Alpha told me so. =P I have no idea how to prove it though. The ratio test results in 1 so that's inconclusive, so I'm left with comparison or limit comparison test..but I don't know what to compare it to.

Thanks in advance for your help!

WolframAlpha tells you the answer, you have the solutions. I don't see what's wrong here
 
flyingpig said:
WolframAlpha tells you the answer, you have the solutions. I don't see what's wrong here

Why are you posting here if you have nothing to contribute??
 
Last edited:
Hi Dick and FlyingPig,

Thanks for your responses.

Dick - you are right, the integral does not converge.

FlyingPig, the problem is that WolframAlpha gave me an integer answer when I entered "the summation of 1/ln(n!) from 2 to infinity", so that was misleading.

Anyway thanks so much for helping me solve the problem.
 
This is what WolframAlpha gave me:

NIntegrate::ncvb: "NIntegrate failed to converge to prescribed accuracy after 9 recursive bisections in n near {n} = {2.176*10^6}. NIntegrate obtained 2.1848934935880644` and 0.0050671018999950595` for the integral and error estimates."

and then the integer 5.49193 as output. I guess that was an estimate for a sum that never actually converges.
 
FuzzieLogic said:
This is what WolframAlpha gave me:

NIntegrate::ncvb: "NIntegrate failed to converge to prescribed accuracy after 9 recursive bisections in n near {n} = {2.176*10^6}. NIntegrate obtained 2.1848934935880644` and 0.0050671018999950595` for the integral and error estimates."

and then the integer 5.49193 as output. I guess that was an estimate for a sum that never actually converges.

You are quite right. Wolfram Alpha can be misleading. So can flyingpig. Can you show the series diverges without trusting either one? What's the integral of 1/(n*ln(n))?
 
Last edited:

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