There are a couple of questions you'll want to answer.

1. Can you use logs to describe the \(D(k)\) function? It's an interesting function because it's kind of a step function.
2. Can you start with some basic values of \(p\), and think about comparisons? Note that \(D(k)\geq 1\) for \(k=1,2,3,4,...\)
3. This is almost something that you could use the Integral Test on...too bad we don't meet the continuity requirement for the integral. Unless...?

Here are some more details about the hints above.

1. We know that \(D(k)=\left\lfloor\log(k)+1\right\rfloor\). So our series is really: $$ \sum_{k=1}^\infty \frac{1}{\left\lfloor\log(k)+1\right\rfloor(k^p)}. $$
2. We can use the fact that $$ \log(k) \leq \left\lfloor\log(k)+1\right\rfloor \leq \log(k)+1 $$ and show then that $$ \frac{1}{\log(k)(k^p)}\geq \frac{1}{D(k)k^p} \geq \frac{1}{\left(\log(k)+1\right)k^p} $$ Something to note is that this inequality only really works when \(k\gt 1\), so we'll consider the infinite series starting at \(k=2\) instead. Adding back the term when \(k=1\) in our original series will not change the behavior, and so we can start our series at \(k=2\) to classify whether it converges or diverges.
3. While \(\dfrac{1}{\left\lfloor\log(x)+1\right\rfloor}\) is not continuous on \([2,\infty)\), the functions \(\dfrac{1}{\log(x)(x^p)}\) and \(\dfrac{1}{(\log(x)+1)(k^p)}\) are. It turns out that the integrals $$ \int_{x=2}^\infty \frac{1}{\log(x)x^p}\;dx $$ and $$ \int_{x=2}^\infty \frac{1}{(\log(x)+1)(x^p)}\;dx $$ are fun to evaluate.

Here are some general thoughts or hints that I might give students.

1. Can you start by comparing the size of \(P_k\) with \(k\)? Which one is bigger? Does this ordering remain for all values of \(k\)?
2. It would be helpful to have a written-out function rule for \(P_k\), but unfortunately we don't have one. Take a look at the Prime Number Theorem in the references at the end of the page. This could be helpful!

Here are some more details about those hints above!

1. Once we convince ourselves that \(P_k\gt k\) for all \(k=1,2,3,...\), then we can get $$ \frac{1}{P_k(k^p)} \lt \frac{1}{k^{p+1}}. $$ This inequality is a great one to use to show some values of \(p\) that will make the \(\displaystyle\sum_{k=1}^\infty \frac{1}{P_k(k^p)} \) converge.
2. There's a lot of fun reading about the Prime Number Theorem, but one of the statements of it gives us a way of approximating prime numbers: $$ P_k \approx k\log(k). $$ We also know that as \(k\to\infty\), \(P_k \to k\log(k)\). This is perfect for the purposes of a Limit Comparison Test! We can use it to compare the two series (again, starting at \(k=2\) to avoid division badness with the logarithm): $$ \sum_{k=2}^\infty \frac{1}{P_k(k^p)} $$ compared to $$ \sum_{k=2}^\infty \frac{1}{k^{p+1}\log(k)}. $$ Hopefully we can pair this with something like the Integral Test to tell us about the behavior of $$ \displaystyle\sum_{k=2}^\infty \frac{1}{k^{p+1}\log(k)} $$ for different values of \(p\) by thinking about the behavior of the integral $$ \int_{x=2}^\infty \frac{1}{x^{p+1}\log(x)}\;dx $$ for different values of \(p\).