Infinite Series Comparison Tests

[Marcivo]

Hello, folks. This happens to be my first post here, and I've come with a question from a problem set in my textbook.

Homework Statement

Determine whether the following series converges or diverges. Give reasons for your answer.

Homework Equations

\sum^{\infty}_{n=2} \frac{1}{ln(n)^2}

The Attempt at a Solution

Now, I've taken a number of approaches toward this problem, but, so far, none of them has provided even an incorrect solution. The integral test fails automatically because the expression is nonintegrable; both the ratio and root tests fail because the solution is equal to 1; I've compared it to \frac{1}{ln(n)}, and \frac{1}{n^2}, as my b_n's, but both are inconclusive by the limit comparison test. The one thing I can think of that would prove the series to be divergent is to say that 1/ln(x)^2 is greater than 1/x, but is this a safe assumption to make?

I went to my professor today, but was just turned away after being told to "apply integration by parts" to the integral of 1/ln(x)^2...which, of course, doesn't work. So, my question is, does anyone here suggest a more solid comparison than 1/x, and, if not, how could I prove that 1/x \leq 1/ln(x)^p for all p \geq 2?

Thanks.

 

[Dick]

Sure. Compare with 1/x. You don't have to assume ln(x)^2<x for sufficiently large x. You can prove it. Look at the limit of ln(x)^2/x as x->infinity. Use l'Hopital.

 

[Marcivo]

Ahh, wonderful. I wrote that down earlier, but I wasn't sure if it would constitute proof to my professor. Thanks for your help.

 

[Dick]

Ahh, wonderful. I wrote that down earlier, but I wasn't sure if it would constitute proof to my professor. Thanks for your help.

The thing is that for the comparison test, you can leave out any finite number of terms in the series. So it only counts what the relation between the two series is in the large n limit.