Question

Does `sum_(n=2)^oo1/(nln(n))` converges ?

Answer 1

The series diverges.

Explanation:

Use the integral test. If we let `u =lnn`, then `du = 1/ndn` and `ndu = dn`

`I = int_2^oo 1/(nlnn) dn`

`I = int_2^oo 1/(n u) * ndu`

`I = int_2^oo 1/u du`

`I = [lnu]_2^oo`

`I = [ln(lnn)]_2^oo`

The limit `lim_(x->oo) ln(lnn)` clearly equals `oo`, therefore the series diverges.

Hopefully this helps!