Determine whether the series is convergent

[azatkgz]

Homework Statement

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}

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.

 

[azatkgz]

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

 

[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.

 

[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}.

 

[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.