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

1 Answer
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>=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)^oo 1/(nlnn)

which is easier to determine using the integral test.

We take as test function:

f(x) = 1/(xlnx)

and as f(x) in the interval x in [2,+oo) is positive and strictly decreasing, and we have:

lim_(x->oo) 1/(xlnx) = 0

and:

f(n) = 1/(nlnn)

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

int_2^oo (dx)/(xlnx)

is also convergent.

Now we have:

int_2^oo (dx)/(xlnx) = int_2^oo (d(lnx))/lnx = [lnabs((lnx))]_2^oo= +oo

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