Twin primes : prove the convergence of $ \lim_{N\sim\infty} \frac{1}{N} \sum^{N}_{p\in T} (\log(p)+\frac{1}{p})²$ Let $T$ be the twin primes set : $p \in T $ if and only if $ p$ and $p+2$ are primes.
Can you help me establish the convergence of:
$$ \lim_{N\sim\infty} \frac{1}{N} \sum^{N}_{p\in T} (\log(p)+\frac{1}{p})^2$$
Is it known that ? :
$$ \lim_{N\sim\infty} \frac{1}{N} \sum^{N}_{p\in T} (\log(p)+\frac{1}{p})^2=B_2$$
Where $B_2$ is the twin primes constant. Can it be proved or is it linked to the conjecture ? (The limit being equal to $B_2$ is a guess based on computational data.)
If you have any other candidate value for the limit.
 A: The limit should be $B_2$, but we can't prove that.
First of all, the difference between $\sum (\log p+1/p)^2$ and $\sum \log^2p$ (where all sums are over the twin primes up to $N$) is less than $3\sum (\log p)/p$, which is $O(\log N)$. So for the limit in question, one might as well look at $\sum \log^2p$.
That sum is equal, by partial summation, to
$$
\sum\log^2p = \int_2^N (\log^2x) \,dT(x) = T(N) \log^2 N - \int_2^N \frac{2T(x)\log x}x\,dx,
$$
where $T(x)$ is the number of twin primes up to $x$. (The first integral is a Riemann-Stieltjes integral, but you can prove that the left-hand sum equals the right-expression directly if you wish.)
If the Hardy-Littlewood conjecture is true, then $T(x) \sim B_2x/\log^2x$, and so
$$
\sum\log^2p \sim B_2N - \int_2^N \frac{2B_2}{\log x}\,dx = B_2N + O\bigg( \frac N{\log N} \bigg),
$$
which indeed implies that your limit equals $B_2$.
But one can also start from your limit equaling $B_2$ and use partial summation in the other direction to deduce the Hardy-Littlewood conjecture for twin primes. So they really are equivalent to each other (and of course, stronger than the qualitative twin primes conjecture itself).
