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

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Homework Statement


\sum_{n=1}^{\infty}\frac{(-1)^n}{\ln(n+1)}


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 \sum_{n=1}^{\infty}\frac{1}{\ln(n+1)}
then using the nth term test/divergence test the sequence a_n=\frac{1}{\ln(n+1)} 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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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.
 
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!
 
Also, try the alternating series test.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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