Let $p_n$ be the $n$-th prime and let $q_1, q_2, \ldots, q_n$ be any rearrangement of the first $n$ primes. Using the rearrangement inequality and the solution to this problem, we can prove that
$$ 1 \le \lim_{n \to \infty}\frac{p_1 + 2p_2 + 3p_3 + \cdots + np_n}{q_1 + 2q_2 + 3q_3 + \cdots + nq_n} \le 2 $$
I was wondering if we can make the above ratio arbitrarily close to any real number in the interval $(1,2)$ by choosing a suitable rearrangement of the first $n$ primes i.e.
Question: Let $\alpha, 1 \le \alpha \le 2$ be any real. Does there always exist a rearrangement such that,
$$ \lim_{n \to \infty}\frac{p_1 + 2p_2 + 3p_3 + \cdots + np_n}{q_1 + 2q_2 + 3q_3 + \cdots + nq_n} = \alpha$$
We have:
The idea is to define parameter $t$, such that the limit takes values in $[1,2]$ depending on $t$.
Turns out the primes $p_n$ are not special, the problem is equivalent if we take either $n$ or $n\log n$.
The general idea here being that, first $(t\times100)\%$ of terms in denominator are arranged in reverse, compared to the numerator.
Trivially, $f(0,g)=1$, for any $g$, as we have the same numerator and denominator.
We know that $f(1, p_x)=2$ which was shown in your other question you linked.
Interestingly, for $g(x)=x$ and for $g(x)=x\log x$, the limit converges in both cases to:
$$ f(t)=\frac{2}{2-t^3} $$
The first two limits I computed with wolfram alpha (Mathematica):
And we are done since: