Can we rearrange primes to make this ratio converge to any real? 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, 
$$ \lim_{n \to \infty}\frac{p_1 + 2p_2 + 3p_3 + \cdots +
 np_n}{q_1 + 2q_2 + 3q_3 + \cdots + nq_n} = \alpha$$

 A: 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 $n\log n$.


For $t \in[0,1]$ and $g:\mathbb N\to \mathbb R$, define:
$$
f(t, g):=\lim_{n\to\infty}\frac
{\sum_{k=1}^{n}k\cdot g(k)}
{\sum_{k=1}^{n - t n}k\cdot g(k)+\sum_{k= n - t n + 1}^{n}k\cdot g((2-t)n-k+1)}
$$
For $g(x)=p_x$ we have your problem, where $p_x$ is the $x$th prime.

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.


We can estimate $p_x\sim x\log x$ like it was done in the linked
question, and discussed in comments there.

Interestingly, for $g(x)=x$ and for $g(x)=x\log x$, the limit converges in both cases to:
$$
f(t)=\frac{2}{2-t^3}
$$

Turns out $f(t,x)\sim f(t,x \log x)\sim f(t, p_x)$. Similar case was
with one of your other
questions.

The first two limits I computed with wolfram alpha (Mathematica):
ClearAll[g, f, t, n, k];
g[n_] := n Log[n];
f[t_] := Limit[Sum[k*g[k], {k, 1, n}]/(Sum[k*g[k], {k, 1, n - t n}] + Sum[k*g[(n - k + n - t n + 1)], {k, n - t n + 1, n}]), n -> Infinity];
f[t]


And we are done since:

*

*The $f(t,x)=f(t, x\log x)=2/(2-t^3)$, depending on $t\in[0,1]$, takes real values in
$[1,2]$

*We can show in a similar way to the question I linked, that $f(t,x)= f(t,x \log x)= f(t, p_x)$
