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I am not sure how to to it and if my approach is correct.

With the integral test I can show that $$\sum_{k=2}^\infty = \frac{1}{k\ln^pk}$$ converges for all $p> 1$, but what about $$\sum_{k=2}^\infty = \frac{1}{k^2\ln^pk}$$ ?

Can I use the comparison test as $\sum_{k=2}^\infty = \frac{1}{k^2\ln^pk} < \sum_{k=2}^\infty = \frac{1}{k\ln^pk}$ and then say that it also converges for all $p > 1$?

Also, how do I find $p$ if I write $ \sqrt{k} $ instead of $k^2 $? For this I am not sure how to start, I would appreciate a hint.

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    It converges for all $p$. Integral test works fine, but the main intuition is that $\log k$ grows so slow that it is eventually beaten by $k^{\epsilon}$ for any $\epsilon > 0$. – 2017-01-15

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For all $p$, it converges due to the Cauchy condensation test:

$$\frac1{n^2\ln^pn}\to\frac{2^n}{(2^n)^2\ln^p2^n}=\frac1{2^nn^p\ln^p2}$$

Which converges for all $p$.

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    Note that $\frac1{n^2\ln^pn}\to0$ for all $p$, so it is eventually a non-increasing sequence. – 2017-01-15
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    @DonAntonio Ok, my bad, it is monotonically decreasing for $n>1$ if $p\ge0$ and $n>e^{-p/2}$ if $p<0$. – 2017-01-15