# Math Help - Test for convergence

1. ## Test for convergence

(1) $\Sigma$(1 to infinity) $(-1)^{n+1} \displaystyle \frac{1-cos(1/n)}{sin(1/n)}$

(2) ) $\Sigma$(2 to infinity) $(-1)^n8^{-n/2}(\displaystyle \frac{n+1}{n})^{n^2}$

(3) $\Sigma$(2 to infinity) $\displaystyle \frac{(-1)^n}{nlnn(lnlnn)^2}$

knowing 3 shud use the cauchy condensation test, how do i apply it, and what test shud i use for the other two?

need to specify if it's absolutely convergent

2. Originally Posted by wopashui
(1) $\Sigma$(1 to infinity) $(-1)^{n+1} \displaystyle \frac{1-cos(1/n)}{sin(1/n)}$
Remembering the identity $\displaystyle \tan \frac{\theta}{2} = \frac{1-\cos \theta}{\sin \theta}$ we 'discover' that the general term is $\displaystyle (-1)^{n+1}\ \tan \frac{1}{2n}$ so that the series converges for the Leibnitz's rule...

Kind regards

$\chi$ $\sigma$