# Determining if the series $\sum_{n=2}^\infty {1\over {nln(n)ln(ln(n))}}$ is convergent or divergent and justifying the answer. [duplicate]

By far, these infinite series and sequences have been my biggest problem in calculus 2, and this problem really puts my limited knowledge to the test.

I have found that posting these pictures helps show everyone what I have learned so far, and then I'll try to solve it myself.

Chapter and section names:

Rules that my book gives me to solve these problems. This shows more of what I should and do know:

Looking at those rules, it doesn't really even seem to help me with this problem. The only way I could see this being solved with what I know is with the integral test.

Also, note that it starts at 2 (I almost missed that myself).

If I'm going to try to use the integral test, then here is my book's definition of it:

And here is the question again: Determining if the series $\sum_{n=2}^\infty {1\over {nln(n)ln(ln(n))}}$ is convergent or divergent and justifying the answer. I honestly don't know where to start with this one, I would try but there are a lot of changes from previous questions, and it makes it a lot harder.

Any help is greatly appreciated. Thank you.

## marked as duplicate by Nosrati, Daniel Fischer calculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Oct 28 '17 at 20:57

• Do you know the integral of $\frac{1}{n\ln n\ln (\ln n) }$? – kingW3 Oct 28 '17 at 20:31
• Cauchy condensation test. – user228113 Oct 28 '17 at 20:31
• @G.Sassatelli I have not learned about that yet. – JustHeavy Oct 28 '17 at 20:32
• how about let $x=ln(n)$? – valer Oct 28 '17 at 20:34
• @kingW3 I came up with $ln(\left\lvert {ln(ln(x))} \right\lvert) + C$ – JustHeavy Oct 28 '17 at 20:35

This is the Cauchy condensation test, look at $\sum 2^n a_{2^n}$. This gives $$\sum \frac{1}{n\ln n}$$ and applying it a second time we get $$\sum \frac{1}{n}$$ which diverges, thus the above two series also diverge.