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Absolute convergence

  • Thread starter dobry_den
  • Start date
  • #1
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Homework Statement


[tex]\sum_{n=2}^{\infty} \ln \left(1+\frac{(-1)^n}{n^p}\right)[/tex]

p is a real parameter, determine when the series converges absolutely/non-absolutely

The Attempt at a Solution



I tried to do the limit [tex]\lim_{n\rightarrow \infty} \frac{\ln \left(1+\frac{(-1)^n}{n^p}\right)}{\frac{(-1)^n}{n^p}}[/tex], which is equal to one and this suggests that the series coverges if p is positive (limit comparison test). But then I'm not sure how to determine the absolute/non-absolute convergence. Could you help me please? Thanks very much in advance!
 
Last edited:

Answers and Replies

  • #2
daniel_i_l
Gold Member
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Try to split the series into two parts - the even numbers and the odd ones and see if they both converge.
 
  • #3
115
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i made a mistake in the first post, the limit comparison test is applicable only to non-negative series. then the limit should be like:
[tex]\lim_{n\rightarrow \infty} \frac{\left|\ln \left(1+\frac{(-1)^n}{n^p}\right)\right|}{\frac{1}{n^p}}[/tex]

which is equal to zero when p is positive. However, the lower series converges when p>1. Therefore, the original logarithm series converges absolutely for p>1.

The textbook then says that the series converges non-absolutely also for 1/2 < p <= 1. But I can't prove it - do you have any ideas?
 
Last edited:

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