Is this series convergent or divergent

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SUMMARY

The series in question, defined as the sum of (-1)^(n-1) * ln(n)/n, is debated for convergence or divergence. The alternating series test confirms that the series is convergent due to the behavior of the ln(n)/n term, which approaches zero as n increases. While one participant argued for divergence by comparing it to the divergent series 1/n, the application of the alternating series test provides a definitive conclusion of convergence. Thus, the series converges based on established mathematical principles.

PREREQUISITES
  • Understanding of the Alternating Series Test
  • Familiarity with the Comparison Test
  • Knowledge of logarithmic functions and their limits
  • Basic principles of series convergence and divergence
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  • Study the Alternating Series Test in detail
  • Explore the Comparison Test and its applications
  • Investigate the behavior of logarithmic functions in series
  • Learn about convergence criteria for series in calculus
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Students of calculus, mathematicians analyzing series convergence, and educators teaching series tests will benefit from this discussion.

Windowmaker
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Homework Statement



Me and my friend are debating on wether the follow seris is convergent or divergent. The seris is the sum of (-1)^n-1 * ln(n)/n.

Homework Equations


p test and comparision tests.

And alternating series test

The Attempt at a Solution


My approach to this problem was that the ln(n) portion grows much slower than n portion. So I compared this function to 1/n. I know 1/n is divergent, so I concluded the above function was also divergent. My friend argues that the limit of ln (n)/n is zero and is greater than ln(n+1)/(n+1). So he says its convergent.
 
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What about the (-1)^(n-1) factor? I think you'd better look at the alternating series test.
 
Windowmaker said:

Homework Statement



Me and my friend are debating on wether the follow seris is convergent or divergent. The seris is the sum of (-1)^n-1 * ln(n)/n.

Homework Equations


p test and comparision tests.

And alternating series test

The Attempt at a Solution


My approach to this problem was that the ln(n) portion grows much slower than n portion. So I compared this function to 1/n. I know 1/n is divergent, so I concluded the above function was also divergent. My friend argues that the limit of ln (n)/n is zero and is greater than ln(n+1)/(n+1). So he says its convergent.
What does the alternating series test say in this case? You seem to be ignoring the fact that this is an alternating series.
 
So by using the alternating series test, this seris is convergent?
 
Also, disregarding for the moment that the series is alternating, when you compare ln(n)/n with 1/n, what must happen for you to conclude that Ʃ(ln(n)/n) diverges?
 
They would have to grow at the same rate?
 
Windowmaker said:
They would have to grow at the same rate?
No, that doesn't have anything to do with the comparison test, which is one of the tests that you listed, and are apparently attempting to use.

This test should be defined in your book. Specifically, there are different inequalities that come into play, depending on whether you are comparing to a convergent series or to a divergent series.
 
Im confused. I am going to go youtube this. Have a nice day.
 

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