# How do you test the alternating series Sigma (-1)^n/(ln(lnn)) from n is [3,oo) for convergence?

Jun 8, 2018

The series:

${\sum}_{n = 3}^{\infty} {\left(- 1\right)}^{n} / \ln \left(\ln n\right)$

is convergent.

#### Explanation:

We have that:

${\lim}_{n \to \infty} \frac{1}{\ln} \left(\ln \left(n\right)\right) = 0$

Consider the function:

$f \left(x\right) = \frac{1}{\ln} \left(\ln \left(x\right)\right)$

As:

$f ' \left(x\right) = - \frac{1}{\ln \left(\ln x\right)} ^ 2 \frac{1}{\ln} x \frac{1}{x} < 0$ for $x > 1$

the function is strictly decreasing in $\left[1 , + \infty\right)$ and thus:

$\frac{1}{\ln} \left(\ln \left(n + 1\right)\right) < \frac{1}{\ln} \left(\ln \left(n\right)\right)$

thus the series:

${\sum}_{n = 3}^{\infty} {\left(- 1\right)}^{n} / \ln \left(\ln n\right)$

is convergent based on Leibniz' theorem.