# Determine whether the series is convergent

1. Oct 26, 2007

### azatkgz

1. The problem statement, all variables and given/known data
Determine whether the series $$\sum_{n=2}^{\infty}a_n$$ is absolutely,conditionally convergent or divergent
$$a_n=\frac{(-1)^n}{\sqrt{n}(\frac{2n}{n+1})^\pi}$$

3. The attempt at a solution
from Abel's test.$$c_n=\frac{(-1)^n}{\sqrt{n}}$$is convergent.and

$$b_n=(\frac{2n}{n+1})^\pi}=\frac{2^{\pi}}{(1+\frac{1}{n})^{\pi}}=\frac{2^{\pi}}{1+\frac{\pi}{n}+o(\frac{1}{n^2})}$$.Which has limit $$2^{\pi}$$.So a_n is convergent.

$$|a_n|=\frac{2^{\pi}}{\sqrt{n}(1+\frac{1}{n})^{\pi}}=\frac{2^{\pi}}{\sqrt{n}+\frac{\pi}{\sqrt{n}}+O(\frac{1}{\sqrt{n}n})}$$

I don't know exactly but it seems to me that the last equation is divergent.So a_n is conditionally convergent.

2. Oct 26, 2007

### azatkgz

Sorry,I wrongly had typed the series .
$$a_n=\frac{(-1)^n}{\sqrt{n}}(\frac{2n}{n+1})^\pi}$$

3. Oct 26, 2007

### Dick

I don't think you can apply Abel's test here, since the basic series you are dealing with, 1/sqrt(n) is of the form 1/n^p and doesn't converge by an integral test. I would look for a proof using the alternating series test.

4. Oct 27, 2007

### azatkgz

The reason why I used Abel's Test is:

Abel's Test

Given two sequences {$$a_n$$} and{$$b_n$$},suppose that
1.The series
$$\sum_{n=1}^{\infty}a_n$$ is convergent.
2.The sequence $$b_n$$ monotonically converges to some number L
Then the series
$$\sum_{n=1}^{\infty}a_nb_n$$ is convergent.

So I choosed first series $$c_n=\frac{(-1)^n}{\sqrt{n}}$$ as convergent.And $$b_n$$ with limit $$2^{\pi}$$.

5. Oct 27, 2007

### Kummer

Let $$a_n = \frac{(-1)^n}{\sqrt{n}}$$ then $$\sum_{n=1}^{\infty} a_n$$ converges. Let $$b_n = \left( \frac{2n}{n+1} \right)^n$$ then $$b_n \leq b_{n+1}$$ and $$\lim b_n \not = \infty$$. This means $$\sum_{n=1}^{\infty} a_nb_n$$ converges.