# How do you use the integral test to determine if  sum_(n=3)^(oo) 1/(nlnnln(lnn)) is convergent or divergent?

Jan 26, 2018

This sum diverges.

#### Explanation:

For a series:

${\sum}_{n = N}^{\infty} f \left(n\right)$

If the integral

${\int}_{N}^{\infty} f \left(n\right)$

is finite then the series will converge.

For:

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

we will have the integral:

${\int}_{3}^{\infty} \frac{1}{x \ln x \ln \left(\ln x\right)} \mathrm{dx}$

Consider the substitution:

$u = \ln \left(\ln \left(x\right)\right)$

Which will give us (bu the chain rule):

$\mathrm{du} = \frac{1}{x \ln x} \mathrm{dx}$

We also have that:

$u = \infty$ at $x = \infty$

and:

$u = \ln \left(\ln 3\right)$ at $x = 3$.

So the integral now becomes:

${\int}_{3}^{\infty} \frac{1}{x \ln x \ln \left(\ln x\right)} \mathrm{dx} = {\int}_{3}^{\infty} \frac{1}{x \ln x} \frac{1}{\ln} \left(\ln x\right) \mathrm{dx}$

${\int}_{\ln} {\left(\ln 3\right)}^{\infty} \frac{1}{u} \mathrm{du} = {\left[\ln u\right]}_{\ln} {\left(\ln 3\right)}^{\infty}$

Note that:

${\lim}_{u \to \infty} \ln u \to \infty$

So evaluating these limits will give:

$\infty - \ln \left(\ln \left(3\right)\right) = \infty$

So the integral does not have a finite values hence the sum diverges.