I was toying around with some prime number related series (trying to generalize some results from a puzzle) and came across this one:
$$\sum_{p \text{ prime}} \frac{1}{p^2+p}$$
Is there any reasonable way to calculate this? If not, can you get some nice bounds on it? (above and below would be nice. I know it lies between $0.3$ and $0.4$, and running it through a program actually tells me it is about $0.33$)
I can't give you an answer, but.
If we consider $$ \sum_{p \leq x} \frac{1}{p(p + 1)} = \sum_{p \leq x} \left( \frac{1}{p} - \frac{1}{p + 1} \right), $$ the question amounts to determining the constant $A$ appearing in the asymptotic formula of the form $\sim \log \log x + A + o(1)$ for the sum $$ \sum_{p \leq x}\frac{1}{p + 1}. $$
I don't think A takes an easy form.