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Determine the values of $p$ for which the given series converges: $$ \sum_{k=3}^\infty\frac1{k\ln k[\ln(\ln k)]^p}. $$

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  • $\begingroup$ what convergence test is best here $\endgroup$ – Integrand Sep 16 at 16:33
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    $\begingroup$ Use the integral test. $\endgroup$ – alex.jordan Sep 16 at 16:36
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My answer may or may not be accurate. It's been a while since I've done power series stuff. I would recommend starting with the integral test because notice that the derivative of $\ln(\ln k)$ is $\frac{1}{k\ln(k)}$. So set up the integral as $$\int_{k=3}^\infty\frac1{k\ln k[\ln(\ln k)]^p}dk$$ Then perform a u substitution: $u=\ln(\ln k)$ and $du=\frac{dk}{k\ln(k)}$ so we have $$\int_{u=\ln(\ln(3))}^\infty\frac{1}{u^p}du$$ This is basically equivalent to a p series, which only converges if p is greater than 1. Perhaps this is how one is supposed to find the answer? Again, I am a bit rusty, and I'm sorry if this is not accurate.

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  • $\begingroup$ Integrale should go from $\ln\ln3$ to $+\infty$ $\endgroup$ – enzotib Sep 16 at 17:15
  • $\begingroup$ Thanks, I fixed it. $\endgroup$ – A Dude Sep 16 at 17:21

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