# Absolute convergence

## 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!

Last edited:

## Answers and Replies

Related Calculus and Beyond Homework Help News on Phys.org
daniel_i_l
Gold Member
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?

Last edited: