Series E from 2 to ~ of 1/n(ln n)^3 diverge or converge?

Question

and Prove too..

I think converge.. is it using integral test..?
help

Answer 1

The integral would be fine; perfect, actually.

Let the corresponding function be

f(x) = 1/x[ln(x)]^3

Now, we want to find

Integral (1/x[ln(x)]^3) dx, {from 2 to infinity}

We solve this integral by substitution.

Let u = ln(x). Then
du = (1/x) dx, so our integral is then

Integral ( [1/(u^3)] du )

Integrating this gives us

(-2/u^2)

Which, after substituting u = ln(x), becomes

[-2/ln(x)]

Since we have an improper integral, we don't want to calculate this from 2 to infinity, but we want to take the LIMIT as x approaches infinity and take the integral of this from 2 to t.

lim [-2/ln(x)] {evaluated from 2 to t}
t -> infinity

lim [ (-2/ln(t)) - (-2/ln(2)) ]
t -> infinity

As t gets very large, (-2/ln(t)) becomes 0, so we have

0 - [-2/ln(2)], OR

2/ln(2)

Which shows that this integral converges.

Therefore, if this integral converges, it follows that the corresponding series will converge too.

Answer 2

assume x=ln(n), then n=e^x, so there is An = 1/(x^3 ·e^x) = e^(-x)/x^3;
take Bn = e^(-x) and Σe^(-x) always converges! Since An