Creating Bounds for $\sum_{k=1}^\infty \frac{1}{p_{p_k}}$ where $p_k$ is the $k$th prime

Question

In another post I asked if $S=\sum_{k=1}^\infty \frac{1}{p_{p_k}}$ (where $p_k$ is the $k$th prime) is irrational, transcendental, or both. I thought that this number was less than $1$ by comparing it to $\sum_{k=1}^\infty \frac{1}{2^k}$ but it turns out that $1/p_{p_k}\sim1/k(\log k)^2$ is greater than $1/2^k$ and the first few terms don't help in creating an upper bound. However I now know that $S>1$ which creates a lower bound. How could I bound $S$ from above? I have a feeling it is less than $2$.

Answer 1

See Rosser's theorem and theories about bounding primes that are related. It is known that $$n\ln n<p_n$$ for $n\ge 1$ so from that (due to monotonicity of the mapping $[x\mapsto\ln x]$) follows $$n\ln n \ln (n\ln n)\le p_n\ln p_n\le p_{p_n}.$$ Therefore $$\sum\frac{1}{p_{p_n}}\le \sum\frac{1}{n\ln n \ln (n\ln n)}$$ at least after some $n$-time like for $n\ge 3$. Sorry for being laconic but I have no time now yet I wanted give some hints how to obtain precise upper bound. Later on we have $$\sum_{n\ge 3}\frac{1}{p_{p_n}}\le \sum_{n\ge 3}\frac{1}{n\ln n \ln (n\ln n)}\le \sum_{n\ge 3}\frac{1}{n\ln^2n }\le \frac{1}{3\ln^2 3}+\int_{3}^{\infty}\frac{\mathrm{d}x}{x\ln^2 x}=\frac{1}{3\ln^2 3}+\frac{1}{\ln 3}.$$

See also Integral test for convergence for the last inequality.

Really fun fact that i discovered. Wolfram Alpha neither Mathematica 12 can not properly evaluated the series $$\sum_{n\ge 3}\frac{1}{n\ln n \ln (n\ln n)}.$$
You can see it here: $\sum_{n\ge 3}\frac{1}{n\ln n \ln (n\ln n)}$ by Wolfram.

Answer 2

I'm not certain, but what about simple bound: $p_n \approx n \log n \Rightarrow p_{p_n} \approx p_{n \log n} \approx n \log n \log (n \log n)$.

Hence we have: $$ \sum_{k} \frac{1}{p_{p_k}} \approx \int_A^{\infty} \dfrac{dx}{x \log x \log (x \log x)} $$

Answer 3

Too long for a comment

Starting from @Marek Kryspin's answer and using Mathematica (version 13.1)

$$S_p=\sum_{n=3}^{10^p}\frac{1}{n \log (n) \log (n \log (n))}$$ gives $$\left( \begin{array}{cc} p & S_p \\ 1 & 0.538719 \\ 2 & 0.691952 \\ 3 & 0.747124 \\ 4 & 0.775812 \\ 5 & 0.793509 \\ 6 & 0.805559 \\ \cdots & \cdots \\ \infty & 0.870730 \\ \end{array} \right)$$

Using the values for $1\leq p\leq 6$ a quick and dirty regression $$S_p=\frac {a+b\,p}{1+c\,p}$$ gives a quite good fit $$\begin{array}{l|lll} \text{} & \text{Estimate} & \text{Std Error} & \text{Confidence Interval} \\ \hline a & -1.80803 & 0.08595 & \{-2.08156,-1.53449\} \\ b & +6.20879 & 0.23007 & \{+5.47661,+6.94098\} \\ c & +7.16857 & 0.26878 & \{+6.31319,+8.02394\} \\ \end{array}$$ and then an asymptotic value of $0.866114$ which does not look totally stupid. The residuals are $$\{-0.000025,0.000196,-0.000169,-0.000185,-0.000022,0.000206\}$$

Concerning $$\int_{3}^\infty \frac{dn}{n \log (n) \log (n \log (n))}=0.720329$$