Question

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

Answer 1

`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.