Does this function have triple log behavior?

Question

Since the Zeta function is related to the primes, I thought of replacing the denominator of $\zeta (s)=\sum^\infty_{k=1} \frac{1}{k^s}$ with the $k$th prime. Turns out this already exists and is called the Prime Zeta function and is denoted by $P(n)$.

$P(1)$ is divergent, in a similar way to $\log \log n$, and the limit of the difference of the two ($P_n(1)-\log \log n$, where $n$ is the number of terms in the sum) approaches a number called Merten's constant $M$. This is analogous to the limiting difference of the harmonic series and the logarithm, which is the number $\gamma$.

So I was thinking that, instead of $\sum^\infty_{k=1} \frac{1}{p_k^s}$ how about we have $\sum^\infty_{k=1} \frac{1}{p_{p_k}^s}$ or the sum of the reciprocals of the $k$th primeth primes? Setting $s=1$ we get a sum analogous to the harmonic series. It can be guessed that this sum approaches triple log behavior analogous the prime harmonic and harmonic series. How can I prove that this is the case (and for all possible number of subscripts) if this guess is correct? Is there a way to graph the two functions together so they can be comparable?

Answer 1

Since $p_k\sim k\log k$, it follows that $p_{p_k}\sim p_k\log p_k\sim k(\log k)^2$. By the limit comparison test, the convergence of $\sum_{k=1}^\infty \frac1{p_{p_k}}$ is the same as that of $\sum_{k=2}^\infty \frac1{k(\log k)^2}$; and this latter series is easily seen to converge by the integral test (so in particular there is no triple-log behavior). Further iterations will of course now converge as well.

Answer 2

I found an interesting convergence test called the Ermakoff test which states that if $$\Bigg |\lim_{x\to\infty}\frac{e^xf(e^x)}{f(x)}\Bigg|<1$$ then $\Sigma f(x)$ converges. Substituting in $p_{p_x}$: $$\Bigg |\lim_{x\to\infty}\frac{e^x}{p_{p_x}p_{p_{e^x}}}\Bigg|\space? \space1$$ Since $p_x > x$ for all $x$ then $p_{p_{e^x}}>p_{e^x}>e^x$ so the limit approaches zero. Hence the sum converges.