TITLE: What is the value of $\sum\limits_{i=1}^\infty\frac{1}{p_{p_i}}$ where $p_{i}$ is the $i$th prime?
QUESTION [18 upvotes]: What is the value of $\sum\limits_{i=1}^\infty\dfrac{1}{p_{p_i}}$ where $p_n$ is the nth prime (and so $p_{p_n}$ is the $k$th prime, where $k$ is the $n$th prime) ?
Thus $\frac{1}{3}+\frac{1}{5}+\frac{1}{11}+...=$ ??
How do I efficiently decide if $\sum_{i=1}^\infty\dfrac{1}{p_{p_i}}\gt1$ is true without using a computer ?
I assume this infinite sum is an irrational number. Has this already been proven ?

REPLY [13 votes]: By the Prime Number Theorem, $p_n\sim n\log n$ to the first approximation. Thus, $p_{p_n}\sim n\log^2 n$ to the first approximation. Since $$\int \frac{dx}{x\log^2 x}=\frac{1}{\log n}$$
the series $\sum \frac{1}{p_{p_n}}$ converges.
In terms of estimating the sum: I know you don't want to use a computer, but I just wanted to point out that I obtained $$\sum_{n=1}^{10^{14}} \frac{1}{p_{p_n}}>1.004$$
by writing a small program in Sage that ran in about a minute. I can post the code if people are interested. Basically it chunks up the sum in increasingly longer intervals and uses the upper bound $p_n<n(\log n + \log(\log n))$.