# How do you use the integral test to determine whether int dx/lnx converges or diverges from [2,oo)?

Jun 25, 2018

The integral

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

is divergent.

#### Explanation:

Note that in the interval $x \in \left[2 , \infty\right)$ the function:

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

is:

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

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

3) Decreasing. In fact $f ' \left(x\right) = - \frac{1}{x {\ln}^{2} x} < 0$

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

So, based on the integral test, the convergence of the integral:

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

is equivalent to the convergence of the series:

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

Now we can easily demonstrate that:

$\ln n < n$

so that:

$\frac{1}{\ln} n > \frac{1}{n}$

and as we know that the harmonic series:

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

is divergent, we can conclude that:

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

is divergent by direct comparison, and hence also:

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

is divergent.