Convergence of a series, lots of logs

  • #1

Homework Statement


Show that this sum is convergent if and only if k>1 :

[tex]\sum_{n=1}^{\infty}\frac{1}{n\log{n}\left(\log{\log{n}}\right)^k}[/tex]


The Attempt at a Solution


I've applied the Cauchy condensation test, giving:

[tex]\frac{2^n}{2^n\log{2^n}\left(\log{\log{2^n}}\right)^k}[/tex]
[tex]=\frac{1}{n\log{2}\left(\log{n\log{2}}\right)^k}[/tex]

Then I figured I just apply it again?

[tex]\frac{2^n}{2^n\log{2}\left(\log{\left(2^n\log{2}\right)}\right)^k}[/tex]
[tex]=\frac{1}{\log{2}\left(n\log{2}+\log{\log{2}}\right)^k}[/tex]

But then what? I'm aiming for something along the lines of:

[tex]\frac{1}{C\cdot n^k}[/tex]

So then the answer is obvious but the best I can come up with now is that the term inside the brackets will come out like:

[tex]\left(n\log{2}+\log{\log{2}}\right)^k = n^k\log{2}^k + \mathit{terms\ of\ order\ <k}[/tex]

And therefore the n^k term will dominate the convergence behaviour... Am I on the right track here or have I missed some algebraic trick? Am I wrong to apply the test a second time? I don't want an answer here if thats possible, just some kind of hint. Thanks in advance.

PS for some reason my tex has failed somewhere but you should get the idea...
 
Last edited:

Answers and Replies

  • #2
benorin
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Intergral Test.
 

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