Does (-1)^(n)ln(n)/n Converge Conditionally?

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SUMMARY

The discussion focuses on the convergence of the series (-1)^(n)ln(n)/n, specifically exploring its conditional convergence using the Alternating Series Test. Participants confirm that the limit of ln(n)/n approaches zero as n approaches infinity, which is a key condition for the test. They establish that ln(x)/x is a decreasing function for large x by analyzing its derivative, (1 - ln(x))/x², and conclude that the series converges conditionally as the terms decrease in absolute value.

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ln(n+1)/(n+1) smaller than ln(n)/(n)

I am trying to show that (-1)^(n)ln(n)/n converges conditionally...
Thank you.

P.s the summation is from 1 to infinity.
 
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Is log(x)/x a decreasing function?
 
frasifrasi said:
is

ln(n+1)/(n+1) smaller than ln(n)/(n)

I am trying to show that (-1)^(n)ln(n)/n converges conditionally...
Thank you.

P.s the summation is from 1 to infinity.

You might find it easier, since you're looking at the Alternating Series Test, to just check the limit as n--> infinity of ln(n)/n . L'Hopital's Rule tells you at once that it's zero.

I've done this one myself and I'm not sure there's a nice way to see the inequality you noted. (And the Ratio Test is unhelpful once again...)
 
SO how would I do it?

I have established that the abs value diverges... so, I have to use something else to prove it's conditionally (this series is supposed to converge conditionally). How can I do this?
 
Last edited:
Prove ln(x)/x is a decreasing function for large x. Take the derivative and see if it's negative.
 
dick, that sounds right.

I got (1 - ln(x))/x^(2)

-- how do I go about showing that this is negative?
 
It would be great if you could show me how to get there. I know I am supposed to figure out on my own by my test is tomorrow and I am trying to learn as much as I can. I want to make sure i know this series before the test.
 
I suppose that the ln(x) will outgrow that 1, so the numerator will eventually become negative, so the series is decreasing and the alternating test says it converges. Is this right?

-- Also, can you take a look at the Urgent post I just made ? : )
 
Figure out when it's zero or undefined. Those are the only places it can change sign. Then test the intervals in between. This is the same old problem of trying to figure intervals where a function is increasing and decreasing. You must have done some of this. You don't even need to be that detailed. What can you say about the derivative if x>e?
 
  • #10
frasifrasi said:
I suppose that the ln(x) will outgrow that 1, so the numerator will eventually become negative, so the series is decreasing and the alternating test says it converges. Is this right?

-- Also, can you take a look at the Urgent post I just made ? : )

I think I already did. Unless you posted something even more urgent.
 
  • #11
frasifrasi said:
I suppose that the ln(x) will outgrow that 1, so the numerator will eventually become negative, so the series is decreasing and the alternating test says it converges. Is this right?

-- Also, can you take a look at the Urgent post I just made ? : )

Yes, that's right.
 

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