Let $p_n$ be the $n$-th prime. Is the following true?
$$ \lim_{n \to \infty}\frac{p_1 + 2p_2 + 3p_3 + \cdots + np_n}{p_n + 2p_{n-1} + 3p_{n-2} + \cdots + np_{1}} = 2 $$
Note: The rearrangement inequality implies that the ratio is $\ge 1$.
Update: Extending Yves Daoust's solution, we can show that following hold
$$ \lim_{n \to \infty}\frac{p_1 + 2^a p_2 + 3^a p_3 + \cdots + n^a p_n}{p_n + 2^a p_{n-1} + 3^a p_{n-2} + \cdots + n^a p_{1}} = a + 1 $$
$$ \lim_{n \to \infty}\frac{p_1^a + 2p_2^a + 3p_3^a + \cdots + np_n^a}{p_n^a + 2p_{n-1}^a + 3p_{n-2}^a + \cdots + np_{1}^a} = a + 1 $$
From the Prime Number theorem,
$$p_n\sim n\log n.$$
Then approximating the sums by integrals,
$$\frac{\int_1^n x^2\log x\,dx}{\int_1^n (n-x)x\log x\,dx}\sim\frac{\frac{n^3}3\log n}{(\frac{n^3}2-\frac{n^3}3)\log n}\to2.$$