# Convergence of a series, lots of logs

## Homework Statement

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

$$\sum_{n=1}^{\infty}\frac{1}{n\log{n}\left(\log{\log{n}}\right)^k}$$

## The Attempt at a Solution

I've applied the Cauchy condensation test, giving:

$$\frac{2^n}{2^n\log{2^n}\left(\log{\log{2^n}}\right)^k}$$
$$=\frac{1}{n\log{2}\left(\log{n\log{2}}\right)^k}$$

Then I figured I just apply it again?

$$\frac{2^n}{2^n\log{2}\left(\log{\left(2^n\log{2}\right)}\right)^k}$$
$$=\frac{1}{\log{2}\left(n\log{2}+\log{\log{2}}\right)^k}$$

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

$$\frac{1}{C\cdot n^k}$$

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:

$$\left(n\log{2}+\log{\log{2}}\right)^k = n^k\log{2}^k + \mathit{terms\ of\ order\ <k}$$

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...

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