# How do you test the series Sigma 1/(ln(n!)) from n is [2,oo) for convergence?

Jan 24, 2017

The series:

sum_(n=2)^oo 1/ln(n!)

is divergent.

#### Explanation:

Given the series:

(1) sum_(n=2)^oo 1/ln(n!)

First we note that for $n \ge 2$:

1/ln(n!) = 1/(lnn + ln(n-1) +...+ln2) > 1/(nlnn)

We can now analyse the convergence of the series:

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

which is easier to determine using the integral test.

We take as test function:

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

and as $f \left(x\right)$ in the interval $x \in \left[2 , + \infty\right)$ is positive and strictly decreasing, and we have:

${\lim}_{x \to \infty} \frac{1}{x \ln x} = 0$

and:

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

all the hypothes of the integral test are satisfied and the series (2) is convergent only if the integral:

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

is also convergent.

Now we have:

${\int}_{2}^{\infty} \frac{\mathrm{dx}}{x \ln x} = {\int}_{2}^{\infty} \frac{d \left(\ln x\right)}{\ln} x = {\left[\ln \left\mid \left(\ln x\right) \right\mid\right]}_{2}^{\infty} = + \infty$

So the series (2) is divergent and then also the series (1) is divergent.