What is the value of $\sum\limits_{i=1}^\infty\frac{1}{p_{p_i}}$ where $p_{i}$ is the $i$th prime? 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 ?
 A: This is a little big for a comment, hence posting it as an answer. Just to summarize everything I could find, 
Wikipedia's super prime article refers to a result by Broughan and Barnett which can then be used to show that the set of super primes is small meaning that the sum of the inverse of its elements converges.
On OEIS sequence of super primes, we see that in one of the comments that
$$\sum_{n>N} \frac{1}{a_n} < \frac{1}{\log(N)}$$
where $a_n$ is the $n$-th super prime, along with the note that this can be shown by the integral test. This doesn't directly answer your question because like daniel said, the error is too large to accurately estimate if the sum is larger or smaller than one. That comment on OEIS was made by Jonathan Sondow who is well established in the field of number theory, geometry, and topology. You might want to dig around his publications to see if he has dealt with super primes anywhere. Otherwise, I would email him.
The value of the sum first exceeds 1 with term 148189304, the reciprocal of the 3081648379th prime (73898684653).
A: 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 $p_n<n(\log n + \log(\log n))$.
