A symbolic combination of an inequality concerning convergent series and Brun's theorem: improving the upper bound

Question

We define from the sequence of twin primes, see this MathWorld the following sequence $$t_n:=\sum_{\substack{1\leq k\leq n\\p_k\in\mathcal{T}}}p_k\tag{1}$$

where we denote the set of all twin primes as $$\mathcal{T}=\{\text{primes }p_k\text{ such that }2+p_k\text{ also is a prime number}\}.\tag{2}$$

Thus $t_n$ is summatory of the first primes $p_k$ belonging to the $k$th pair of primes $(p_k,p_k+2)$. Here for $n\geq 1$ we consider that if there exist finitely many twin primes, our sequence $(1)$ will be finite (i.e our sequence $t_n$ isn't an infinite sequence, in the case that there are finitely many twin primes our sequence $t_n$ has a last term and thus is a finite sequence), and the subscript $n$ runs from $1$ to the index of the last twin prime.

Then using (symbolically) the so-called Brun's theorem and an inequality for series that I found from the literatute (see Exercise 215, page 109 of [1]) one gets (unconditionally) the following Claim.

Claim. For every natural $n$ $$\sum_{1\leq k\leq n}(2k+1)\left(\frac{1}{t_k}+\frac{1}{2k+t_k}\right)<4B,\tag{3}$$ where $B$ denotes the Brun's constant. Thus $$\sum_{1\leq k,\text{ } p_k\in\mathcal{T}}(2k+1)\left(\frac{1}{t_k}+\frac{1}{2k+t_k}\right)<4B.\tag{4}$$

I would like to know if it is possible to improve the upper bound in $(2)$, that is I would like to know if my combination of Brun's theorem and the cited inequality has a good result, or can be improved with standard calculations (feel free if you want to add more elaborated calculations).

Question. Can you justify with a proof that we've an improvement of the form $$\sum_{1\leq k,\text{ } p_k\in\mathcal{T}}(2k+1)\left(\frac{1}{t_k}+\frac{1}{2k+t_k}\right)<\text{ a more tight upper bound }<4B\,?\tag{5}$$ If you are able to justify an improvement of our upper bound in $(4)$ on assumption of some remarkable conjectures, feel free to answer also from this way. Many thanks.

I believe that this kind of questions were not in the literature (this kind of combinations of inequalities for convergent/divergent series and twin primes).

We know the asymptotic of the summatory $$\sum_{\text{twin primes }p,p+2\leq x}p\tag{6}$$ from an answer of one of my previous post [2].

References:

[1] Jack D’Aurizio, Superior Mathematics from an Elementary point of view, course notes, University of Pisa (2017-2018).

[2] A conditional asymptotic for $\sum_{\text{$p,p+2$ twin primes}}p^{\alpha}$, when $\alpha>-1$, this Mathematics Stack Exchange (2015).

Answer 1

Not exactly what you want, but it can be a starting point. It is not difficult to see that $$t_{k}=\sum_{m\leq k,\,p_{m}\in\mathcal{T}}p_{m}=\sum_{m\leq k,\,p_{m}\in\mathcal{T}}\frac{p_{m}\left(2p_{m}+2\right)}{2p_{m}+2}>\sum_{m\leq k,\,p_{m}\in\mathcal{T}}\frac{p_{m}\left(p_{m}+2\right)}{2p_{m}+2}$$ then, by the D'Aurizio exercise, we have $$\sum_{k\leq n}\left(2k+1\right)\left(\frac{1}{t_{k}}+\frac{1}{t_{k}+2k}\right)<2\sum_{k\leq n}\frac{2k+1}{t_{k}}$$ $$<8\sum_{k\leq n,\,p_{k}\in\mathcal{T}}\frac{2p_{k}+2}{p_{k}\left(p_{k}+2\right)}<8\sum_{k\geq1,p_{k}\in\mathcal{T}}\left(\frac{1}{p_{k}}+\frac{1}{p_{k}+2}\right)=8B.$$ Maybe the result can be improved but at this moment I don't see a good way to fix it.