# Sum of the reciprocal of the prime-position primes.

The primes are $2, 3, 5, 7, 11, 13...$

The sum of the reciprocals of the primes diverges, proven by Euler:

$$\sum_{n=1}^\infty{\frac{1}{p_n}}=\infty$$

Here, $p_n$ is the $n$-th prime.

I'm asked to prove whether the following converges or diverges:

$$\sum_{n=1}^\infty{\frac{1}{p_{p_n}}} = \frac{1}{3} + \frac{1}{5} + \frac{1}{11} + \cdots$$

I think that it diverges, but I don't know how to show it. Is it true that if I take any infinite subset of a diverging series, the subseries will diverge? That doesn't sound true.

$$\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots$$

The above series is a subseries of $(1 + \frac{1}{2} + 1 + \frac{1}{4} + 1 + \frac{1}{8} + \cdots)$. The full series diverges but the subseries converges, and both are infinite in length.

How do I solve this problem?

• For what it's worth, Wolfram Alpha says "the ratio test is inconclusive." I tried Sum[1/Prime[Prime[n]], {n, Infinity}]. – David R. Apr 5 '16 at 21:09
• Related (but not answered) : math.stackexchange.com/questions/1086127 – Watson Apr 6 '16 at 6:49
• Is this series value known with any accuracy? If someone knows a reference, it would be great. If not, I'll torture Mathematica with it tonight – Yuriy S Apr 8 '16 at 14:25
• The value of the series seems to be around $1$. Mathematica gives $\approx 0.9772$ for $100~000$ terms and $\approx 0.9804$ for $200~000$ terms – Yuriy S Apr 8 '16 at 23:28

## 1 Answer

I think it is a duplicate, but in any case, if we have a weak version of the PNT, stating that $\pi(n)\leq\frac{Cn}{\log n}$ (for instance, Chebyshev's bound), we also have $p_n\geq D\,n \log n$ and $p_{p_n}\geq D\, n \log^2 n$ for any $n$ big enough.

The series $\sum_{n\geq 2}\frac{1}{n\log^2 n}$ is convergent by the Cauchy's condensation test,
hence $\sum_{n\geq 1}\frac{1}{p_{p_n}}$ is convergent, too.