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?
As has been pointed out in the comments, $\operatorname{lcm}(p,p+1)=p(p+1)$ for $p$ prime. Contrary to what has been written in the comments, the sum converges since the terms are $O\left(p^{-2}\right)$; the series cannot be split into separate sums using partial fraction decomposition because the terms of the individual sums would be $O\left(p^{-1}\right)$, so these sums don’t converge individually.
We have
\begin{eqnarray} \sum_p\frac1{p(p+1)} &=& \sum_p\frac1{p^2}\cdot\frac1{1+\frac1p} \\ &=& \sum_p\sum_{s=2}(-1)^sp^{-s} \\ &=& \sum_{s=2}(-1)^s\sum_pp^{-s} \\ &=& \sum_{s=2}(-1)^sP(s)\;, \end{eqnarray}
where $P(s)$ is the prime zeta function. Wolfram|Alpha evaluates this sum to approximately $0.33023$.