# Use the integral method to determine if the series converge or diverge:?

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

Jun 18, 2018

The series:

${\sum}_{n = 2}^{\infty} \frac{1}{n \ln n}$

is divergent.

#### Explanation:

Consider the function;

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

As:

(1) $f \left(x\right) > 0$ for $x > 1$

(2) ${\lim}_{x \to \infty} f \left(x\right) = 0$

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

so that $f \left(x\right)$ is strictly decreasing in $\left(1 , + \infty\right)$.

(4) $f \left(n\right) = \frac{1}{n \ln n}$ for $n \ge 2$

Then the convergence of the series:

${\sum}_{n = 2}^{\infty} \frac{1}{n \ln n}$

is equivalent to the convergence of the improper integral:

${\int}_{2}^{\infty} \frac{\mathrm{dx}}{x \ln x}$

Now:

${\int}_{2}^{\infty} \frac{\mathrm{dx}}{x \ln x} = {\lim}_{t \to \infty} {\int}_{2}^{t} \frac{\mathrm{dx}}{x \ln x}$

${\int}_{2}^{\infty} \frac{\mathrm{dx}}{x \ln x} = {\lim}_{t \to \infty} {\int}_{2}^{t} \frac{d \left(\ln x\right)}{\ln} x$

${\int}_{2}^{\infty} \frac{\mathrm{dx}}{x \ln x} = {\lim}_{t \to \infty} {\left[\ln \left(\ln x\right)\right]}_{2}^{t}$

${\int}_{2}^{\infty} \frac{\mathrm{dx}}{x \ln x} = - \ln \left(\ln 2\right) + {\lim}_{t \to \infty} \left(\ln \left(\ln t\right)\right) = + \infty$

So the series is divergent.