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

In summary: You are quite right. Wolfram Alpha can be misleading. So can flyingpig. Can you show the series diverges without trusting either one?
  • #1
FuzzieLogic
3
0

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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  • #2
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
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
 
  • #4
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??
 
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  • #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
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:

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

1. What is the formula for determining convergence of (-1)^n/ln(n) as n goes to infinity?

The formula for determining convergence of a series is lim n→∞ |an| = 0. In this case, the formula would be lim n→∞ |(-1)^n/ln(n)| = 0.

2. Is (-1)^n/ln(n) a convergent or divergent series?

This series is divergent, as the limit of its absolute value does not equal 0 when n approaches infinity.

3. How can we prove that (-1)^n/ln(n) is divergent?

We can prove that this series is divergent by using the direct comparison test. We can compare it to the harmonic series, 1/n, which is also divergent. As n approaches infinity, the value of ln(n) will increase much faster than n, causing the alternating series to become larger and larger instead of approaching 0.

4. Can we use the alternating series test to determine the convergence of (-1)^n/ln(n) as n goes to infinity?

No, the alternating series test only applies to alternating series where the absolute value of the terms decreases as n increases. In this case, the absolute value of the terms does not decrease, so the alternating series test cannot be used.

5. Are there any other tests we can use to determine the convergence of (-1)^n/ln(n) as n goes to infinity?

Yes, we can also use the ratio test to determine the convergence of this series. The ratio test states that if lim n→∞ |an+1/an| < 1, then the series converges. In this case, the limit would be lim n→∞ |(-1)^(n+1)ln(n+1)/(-1)^nln(n)|, which simplifies to lim n→∞ |ln(n+1)/ln(n)| = 1. Since this limit is equal to 1, the series is inconclusive and we cannot determine its convergence using the ratio test.

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