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

1 Answer
Mar 12, 2017

sum_(n=2)^oo 1/(n ln n)" " diverges.

Explanation:

Note that:

d/(dx) ln x = 1/x

So:

d/(dx) ln(ln x) = 1/(ln x) * d/(dx) ln x

color(white)(d/(dx) ln(ln x)) = 1/(ln x) * 1/x

color(white)(d/(dx) ln(ln x)) = 1/(x ln x)

So:

int 1/(x ln x) dx = ln(ln x) + C

Then:

lim_(x->oo) ln(ln x) = oo

So by the integral test:

sum_(n=2)^oo 1/(n ln n)" " diverges.