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How would you test the following series for convergence?

$$\sum_{n=2}^{\infty}\frac{1}{n(ln(n))^p(ln(ln(n)))^q}$$ where p>1 and q>1?

I tried using integral test by setting $$u=ln(ln(n))$$ and $$du=\frac{1}{n(ln(n))}$$ but because of the p and q values I don't believe this test is possible. I was also thinking along the lines of Comparison Test but was not sure what to compare it to?

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    $\begingroup$ Compare with $\frac{1}{n(\log n)^p}$ which can be shown to converge via integral test; $$\int\frac{dx}{x}\frac{1}{(\log x)^p}=\int \frac{du}{u^p}.$$ $\endgroup$ – blue Apr 24 '14 at 21:29
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$\sum_{n=2}^{\infty}\frac{1}{n(ln(n))^p(ln(ln(n)))^q} < \sum_{n=2}^{\infty}\frac{1}{(ln(n))^p}$, converges for $p>1$, by cauchy-condensation test !

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Use what you know -- you determined convergence in the case when p=1. Now consider a direct comparison to the original series -- certainly for n sufficiently large, the original series is strictly less than the one where p=1.

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  • $\begingroup$ I think you're on the right track here, but a successful comparison requires using $q > 1$, something you probably know but omitted to mention. $\endgroup$ – hardmath Feb 22 '14 at 20:29
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You can also apply the Cauchy condensation test once or twice. If $p=1$, then $q>1$ is necessary for convergence, if $p>1$, the value of $q$ is not that important.

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Hint: Forget about the $\ln\ln$, in the long run it only makes things smaller. Then use the Integral Test. We don't even really need $q\gt 1$.

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