Determine whether the series converges absolutely, converges conditionally or diverges.

[azatkgz]

Homework Statement

Determine whether the series converges absolutely,converges conditionally or diverges.

$$\sum_{n=1}^{\infty}\ln\left(1+\frac{(-1)^n}{n^p}\right)$$
where p is a some parameter

The Attempt at a Solution

$$\ln\left(1+\frac{(-1)^n}{n^p}\right)=\frac{(-1)^n}{n^p}-\frac{1}{n^{2p}}+\frac{(-1)^{3n}}{3n^{3p}}+O(\frac{1}{n^{4p}})$$

Here
$$\sum_{n=1}^{\infty}\frac{1}{n^{2p}}$$ converges for p>1/2

$$\sum_{n=1}^{\infty}\frac{(-1)^n}{n^p}$$ converges absolutely for p>1

My answer is the series converges for p>1/2

for $$\frac{1}{2}<p\leq 1$$ it converges conditionally

for $$p>1$$ it converges absolutely

 

[mighty2000]

.

Thank you for your question. After analyzing the series, I can confirm that your answer is correct. The series converges for p>1/2, converges conditionally for 1/2<p≤1, and converges absolutely for p>1.

To further explain, when p>1/2, the series can be compared to the convergent series \sum_{n=1}^{\infty}\frac{1}{n^{2p}}. This comparison test shows that the given series also converges.

When 1/2<p≤1, the series can be compared to the alternating harmonic series \sum_{n=1}^{\infty}\frac{(-1)^n}{n}. This series is known to converge conditionally, meaning that the series of absolute values \sum_{n=1}^{\infty}\frac{1}{n} also converges, but the original series does not converge absolutely.

Lastly, when p>1, the series can be compared to the convergent series \sum_{n=1}^{\infty}\frac{1}{n^{2p}} and \sum_{n=1}^{\infty}\frac{(-1)^n}{n^p}. Both of these series converge absolutely, so the given series also converges absolutely.

I hope this helps clarify the convergence of the series. Keep up the good work in your studies!