$\sum_{i=1}^{\infty}\frac{1}{p(i)p(i+1)}$, where $p(i)$ is the $i$th prime number

Question

I know that the sum of the reciprocals of the primes diverges, but I was asked of this question (title).

The problem stated that it will converge to a value, but I cannot figure out the value.

Answer 1

Just for the fun of playing with "small" numbers.

Considering $$S_k=\sum_{i=1}^{10^k}\frac{1}{p_i\,p_{i+1}}$$ what I obtained (better say, my computer) is $$\left( \begin{array}{cc} k & S_k \\ 1 & 0.2944055709 \\ 2 & 0.3008401966 \\ 3 & 0.3010804833 \\ 4 & 0.3010924131 \\ 5 & 0.3010931251 \end{array} \right)$$ My computer (and I) gave up for $k=6$.

Inverse symbolic calculators do not seem to find anything.

The upper bound is $$\sum_{i=1}^{\infty}\frac{1}{p_i^2}=P(2)\approx 0.452247$$ where appears the prime zeta function.

Edit

You could be interested by this which gives $\approx 0.301093176358399894$

Answer 2

$\frac{1}{p(i)p(i+1)} < \frac{1}{p(i)^2}$, so $\sum \frac{1}{p(i)p(i+1)} < \sum \frac{1}{p(i)^2}$. On the other hand, $\sum \frac{1}{p(i)^2} < \sum \frac{1}{n^2}$, because the former sum is a subset of the latter. Finally, $\sum \frac{1}{n^2} = \frac{\pi^2}{6}$ is known to converge.