Consider the following summation:
$$\sum_{p \in \mathbb P}^{ }\frac{1}{\operatorname{lcm}\left(p,p+1\right)}$$
Where $\mathbb P$ is the set of prime numbers.
The summation is bounded since:
$$\sum_{p \in \mathbb P}^{ }\frac{1}{\operatorname{lcm}\left(p,p+1\right)}<\sum_{k=2}^{\infty}\frac{1}{\operatorname{lcm}\left(k,k+1\right)}<\sum_{k=2}^{\infty}\frac{1}{\operatorname{lcm}\left(k,k^{2}\right)}=\sum_{k=2}^{\infty}\frac{1}{k^{2}}=\frac{\pi^{2}}{6}-1≈ 0.644934066848$$
Which follows from the Basel problem.
My question is what is the asymptotic behavior of this summation? is there any better upper bound?
$$\sum_{p \in \Bbb P} \frac 1 p - \frac 1 {p+1}$$
The sum $\sum 1 / p$ is known to be asymptotic to $\log \log n$. More explicitly, per this answer,
$$\sum_{p \in \Bbb P \ p \le n} \frac 1 p = \log \log n + \mathcal O(1)$$
[cont.]
– PrincessEev Mar 17 '20 at 17:40$$\lim_{n \to \infty} \left( \sum_{p \in \Bbb P \ p \le n} \frac 1 p - \log \log n \right) = M$$
I'm not 100% sure how this might be useful yet, I'm a bit lost in applying it, but it's a start.
– PrincessEev Mar 17 '20 at 17:40