How do you use the integral test to determine if # sum_(n=3)^(oo) 1/(nlnnln(lnn))# is convergent or divergent?

1 Answer
Jan 26, 2018

This sum diverges.

Explanation:

For a series:

#sum_(n=N)^oof(n)#

If the integral

#int_N^oof(n)#

is finite then the series will converge.

For:

#sum_(n=3)^oo1/(nlnnln(lnn))#

we will have the integral:

#int_3^oo1/(xlnxln(lnx))dx#

Consider the substitution:

#u=ln(ln(x)) #

Which will give us (bu the chain rule):

#du=1/(xlnx)dx#

We also have that:

#u=oo# at #x=oo#

and:

#u=ln(ln3)# at #x=3#.

So the integral now becomes:

#int_3^oo1/(xlnxln(lnx))dx=int_3^oo1/(xlnx)1/ln(lnx)dx#

#int_ln(ln3)^oo1/udu=[lnu]_ln(ln3)^oo#

Note that:

#lim_(u->oo)lnu->oo#

So evaluating these limits will give:

#oo-ln(ln(3))=oo#

So the integral does not have a finite values hence the sum diverges.