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

Question

11187 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-24T22:43:11

The series:

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

is divergent.

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.