Absolute Convergence of Homework Series: Real Parameter p

• dobry_den
In summary, the conversation discusses the convergence of the series \sum_{n=2}^{\infty} \ln \left(1+\frac{(-1)^n}{n^p}\right), where p is a real parameter. The attempt at a solution involves using the limit comparison test to determine the convergence and absolute/non-absolute convergence of the series. It is found that the series converges absolutely for p>1, but the proof for non-absolute convergence for 1/2 < p <= 1 is not provided.

Homework Statement

$$\sum_{n=2}^{\infty} \ln \left(1+\frac{(-1)^n}{n^p}\right)$$

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

The Attempt at a Solution

I tried to do the limit $$\lim_{n\rightarrow \infty} \frac{\ln \left(1+\frac{(-1)^n}{n^p}\right)}{\frac{(-1)^n}{n^p}}$$, 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!

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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.

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:
$$\lim_{n\rightarrow \infty} \frac{\left|\ln \left(1+\frac{(-1)^n}{n^p}\right)\right|}{\frac{1}{n^p}}$$

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?

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