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

  • #1

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!
 

Answers and Replies

  • #2
Dick
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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?
 
  • #3
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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
 
  • #4
Dick
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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:
  • #5
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.
 
  • #6
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.
 
  • #7
Dick
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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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