Stumped on yet another absolute convergence problem. (Converge conditionally, absolutely, or diverges)
$$\sum_{n=1}^\infty \frac{(-1)^n \ln(n)}{n} $$
First, I tested for absolute convergence, and the series reduced to $\frac{\ln(n)}{n}$ .....Using direct comparison, $\frac{\ln(n)}{n} > \frac{1}{n}$ I determined this diverges. So, it does not converge absolutely.
Next, to use the alternating series test on the original series, you need to make sure the magnitude of the terms are decreasing. Well, they are not:
$$\sum_{n=1}^\infty \frac{(-1)^n \ln(n)}{n} = -0 + .35 - .37 + .35.... $$
So, without using the alternating series test, how do I determine if the orig. series converges or diverges?
How can I use the divergence test (Nth term test)? I'm not sure what the nth term tends to with that (-1) oscillation. Just focus on $\frac{\ln(n)}{n}$ But we established that diverges, right? But, I am told this series converges conditionally.
Do I use integral test?