Absolute Convergence of Series with (-1)^n and 1/ln(n+1) Terms

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It may be helpful.In summary, using the absolute convergence test and nth term test, it can be shown that the series \sum_{n=1}^{\infty}\frac{(-1)^n}{\ln(n+1)} converges absolutely. However, the book claims that it only converges conditionally. To prove this, other tests such as the comparison test or integral test may be used. Additionally, the alternating series test may also be helpful in showing convergence.
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miglo
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Homework Statement


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


Homework Equations


absolute convergence test
nth term test/divergence test


The Attempt at a Solution


so the absolute convergence test says that if the absolute value of the series converges then the original series converges absolutely
so with the series i have, in absolute value is [tex]\sum_{n=1}^{\infty}\frac{1}{\ln(n+1)}[/tex]
then using the nth term test/divergence test the sequence [itex]a_n=\frac{1}{\ln(n+1)}[/itex] goes to zero as n goes to infinity therefore the series converges, so i have absolute convergence
but my book says that it only converges conditionally, what am i doing wrong? or is the book wrong?
 
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  • #2
The nth term test can only show you that a series diverges. It can't prove it converges. I'd suggest you try a comparison test or an integral test to show 1/ln(n+1) diverges.
 
  • #3
oohhh right, its been awhile for me since I've done problems on series
cant believe i forgot how the nth term test works, thanks!
 
  • #4
Also, try the alternating series test.
 

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