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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?

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    Well, clearly, $\text{lcm}(p,p+1) = p(p+1)$, so your sum is equivalent to, after partial fraction decomposition,

    $$\sum_{p \in \Bbb P} \frac 1 p - \frac 1 {p+1}$$

    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
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    and, in a vein similar to the harmonic series, there exists a constant $M \approx 0.261497$ such that

    $$\lim_{n \to \infty} \left( \sum_{p \in \Bbb P \ p \le n} \frac 1 p - \log \log n \right) = M$$

    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
  • @EeveeTrainer, WOW, Indeed I did not know about that constant, really thanks. –  Mar 17 '20 at 17:41
  • You have $\operatorname{lcm}(n, n + 1) = n (n + 1) / \gcd(n, n + 1) = n (n + 1)$.... – vonbrand Mar 17 '20 at 20:46
  • A179119 in the OEIS gives the corresponding numerical value. – Varun Vejalla Mar 19 '20 at 03:20

1 Answers1

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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 $0.33023$.

joriki
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