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I know that the series diverge, I'm just having hard time showing it.

$$\sum_{n=3}^\infty\frac1{n(\log n)(\log\log n)}$$

Thanks in advance

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Please learn how to type formulas on the site. – Did May 17 '13 at 19:36

The Cauchy Condensation Test will do it.

If you have not seen it, here is what it says. Suppose that $f(n)$ is positive nonincreasing. Then $\sum_{a}^\infty f(n)$ converges if and only if $\sum_a^\infty 2^nf(2^n)$ converges.

In our case, we have $f(n)=\frac{1}{n\log n\log\log n}$. Thus $2^n f(2^n)=\frac{1}{(n\log 2)(\log n+\log\log 2)}$. The divergence of $\sum 2^n f(2^n)$ now follows from the possibly familiar fact that $\sum_2^\infty \frac{1}{n\log n}$ diverges. If that fact is not familiar, we can use Cauchy Condensation again.

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HINT: Use the integral test. There’s a nice $u$-substitution available to handle the integral.

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thak you, dont know why i didn't try it before... – matan May 17 '13 at 17:51
@matan: You’re welcome. – Brian M. Scott May 17 '13 at 17:58

Hint: Use the comparison test with 1/log(n)

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This is incorrect. If $ \sum \frac{1}{\log n} $ converged, then this sum would converge. However, that sum diverges so it says nothing about the convergence of this sum. – Jon Claus May 17 '13 at 17:55

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