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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 ?

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For a miserable estimate, you could use Bertrand's theorem that says for any integer $n$ there exists at least one prime $p$ with $n<p<2n$. If $p$ is a prime, then the next prime prime, $q$, must occur within $p$ primes later. Thus, $p<q<2^p p$. Thus, the $k$th prime prime $p_k$ must satisfy $\frac{1}{p_k} > 1/a_k$, where $a_1=2$ and $a_{k+1}=2^{a_k} a_k$. This, sadly, won't give you a lower bound of one. –  nayrb Jun 26 '13 at 22:51
    
Actually, I'm not convinced of the statement, " If p is a prime, then the next prime prime, q, must occur within p primes later." I need to think about it. –  nayrb Jun 26 '13 at 22:54
    
A guess is that the series converges a little faster than $\frac{1}{n\log^2 n}$. Rationality/irrationality question may not be accessible with known tools. –  André Nicolas Jun 26 '13 at 22:57
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The $k$-th term in the sum is the $p_k$-th prime?, so $1/p_{p_1} + 1/p_{p_2} + 1/p_{p_3} + 1/p_{p_4} + \cdots$? –  Daniel Fischer Jun 26 '13 at 22:57
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OEIS A006450 and a comment that the sum of reciprocals of the sequence converges, along with a brief explanation of why. A cite too. 'Primes with a prime subscript', Dressler et al., JACM vol 22, issue 3 pp 380-81 (1975) –  daniel Jun 26 '13 at 23:31

2 Answers 2

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 $p_n<n(\log n + \log(\log n))$.

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Wow, I would be very interested in the code if you can paste it here. What is the largest prime number sage can handle on whatever machine you are doing this on? Are you saying that sage can work with Prime[Prime[10^14]]? And what about accuracy and how many digits of computation are you carrying? –  Fixed Point Jun 28 '13 at 7:30
    
I only work with primes up to a certain cutoff. Beyond that cutoff, I simply use the upper bound for $p_n$. –  pre-kidney Jun 28 '13 at 8:06
    
Okay, that's what I figured. That's the problem with the upper bound, it isn't that tight so your result of 1.004 could be too big of an overestimate and the actual sum might still be less than 1. I tried doing this in mathematica carrying a 100 digits with a hundred million terms using primes themselves. It took its sweet time (another reason to be surprised when you said it ran in about a minute) and the result was less than one. Don't quote me but I think it was something like 0.98 when I was just playing around with some of this stuff without any serious effort. –  Fixed Point Jun 28 '13 at 8:15
    
+1 nice work though, even computationally it is quite a challenge. –  Fixed Point Jun 28 '13 at 8:16
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By the way, I had previously computed an upper bound of 1.06, and with a longer computation I found a better lower bound of 1.02. This sum isn't that hard to work with, and I'm sure others have better results as well. –  pre-kidney Jun 28 '13 at 8:22

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).

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