Help with condensation test

**Stormey** · January 21st 2013, 09:50 AM

i guys.
i need to determine if this serie converges:

$\sum_{n=2}^{\infty}\frac{1}{n\ln(n)\ln(\ln(n))}$

this is what i've tried so far:

i use condensation test and got the following:

$\sum \frac{2^n}{2^n\ln2^n[\ln(\ln2^n)]}=\frac{1}{n\ln2[\ln(n\ln2)]}$

$\frac{1}{\ln2}\sum \frac{1}{n[\ln(n\ln2)]}=\frac{1}{n[\ln n+\ln(\ln2)]}=\frac{1}{n\ln n+n\ln(\ln2)}$

$\frac{1}{\ln2}\sum \frac{1}{n\ln n+n\ln(\ln2)}=\frac{1}{n\ln n}+\frac{1}{n\ln(\ln2)}$

now, my question is this:

can i say that since: $\sum \frac{1}{n\ln(\ln2)}$ does not converge the the whole thing does'nt converge?

is there more elegant way to handle this horrible serie?

thanks in advanced!

**chiro** · January 21st 2013, 06:40 PM

Hey Stormey.

Have you come across the Integral Test for convergence? I would suggest looking into that with the appropriate substitution.

**hollywood** · January 21st 2013, 07:12 PM

Yes, chiro is right, the integral test works.

- Hollywood

Here is a link to the condensation test: Cauchy condensation test - Wikipedia, the free encyclopedia

**Stormey** · January 21st 2013, 10:40 PM

hi, and thanks for the help.

we didn't learn the integral test yet, so i'm supposed to answer it without it, i guess...

**hollywood** · January 22nd 2013, 06:39 AM

Maybe this condensation test will do it.

$\sum_{n=2}^{\infty}\frac{1}{n\ln(n)\ln(\ln(n))} =$

which converges if and only if the following series converges:

$\sum_{n=2}^{\infty}\frac{2^n}{2^n\ln(2^n)\ln(\ln(2 ^n))} =$

$\sum_{n=2}^{\infty}\frac{1}{n\ln(2)\ln(n\ln(2))} =$

$\frac{1}{ln(2)}\sum_{n=2}^{\infty}\frac{1}{n(\ln(n )+\ln(\ln(2)))}$

This is what you did in your original post. Now let's do the condensation test again. Our original series converges if and only if the following series converges:

$\frac{1}{ln(2)}\sum_{n=2}^{\infty}\frac{2^n}{2^n(\ ln(2^n)+\ln(\ln(2)))} =$

$\frac{1}{ln(2)}\sum_{n=2}^{\infty}\frac{1}{n\ln(2) +\ln(\ln(2))}$

We think this diverges, so we simplify by making it smaller. We have $\ln(ln(2))\le 2\ln(2)\le n\ln(2)$ , so our new sum is greater than or equal to:

$\frac{1}{ln(2)}\sum_{n=2}^{\infty}\frac{1}{n\ln(2) + n\ln(2)}=$

$\frac{1}{2ln(2)}\sum_{n=2}^{\infty}\frac{1}{n\ln(2 )}=$

$\frac{1}{2(ln(2))^2}\sum_{n=2}^{\infty}\frac{1}{n}$

which diverges.

- Hollywood

**Stormey** · January 22nd 2013, 11:26 AM

brilliant!
thank you very much Hollywood.

didn't know i can use this test twice like that.

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