Question

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

Answer 1

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.