Let $n \in \mathbb N$ and $k_n \in \left\{0,..,n \right\}$ then we define the numbers
$$x_{n,k_n} = \frac{k_n+n^2}{n^3+n^2}.$$
It is easy to see that these numbers satisfy
$$x_{n,0} = \frac{1}{n+1} \le x_{n,k_n} \le x_{n,n} =\frac{1}{n}.$$
I would like to know whether there exist three constants $C_1,C_2,C_3>0$ and an integer $i \in \mathbb N$ such that we can find for every $x_{n,k_n}$ a $reduced$ fraction $$\frac{p_{n,k_n}}{q_{n,k_n}}$$ such that two conditions hold:
1.) The denominator can be controlled nicely:
$$ \frac{C_1}{n^i} \le \frac{1}{q_{n,k_n}} \le \frac{C_2}{n^3}$$ and
2.) The approximation is sufficiently good:
$$\left\vert x_{n,k_n}-\frac{p_{n,k_n}}{q_{n,k_n}} \right\vert \le \frac{C_3}{n^3}.$$
So to summarize: I am wondering whether one can approximate the $x_{n,k_n}$ by reduced fractions up to an error of order $1/n^3$ and whether those fractions can have a denominator that is always between two different powers of $1/n^k.$
I guess we can even keep $q_{n,k_n}=n^3+n^2$. Without this strict restriction let $q_{n,k_n}=q$ be the largest prime number less than $n^3$. R. C. Baker, G. Harman, and J. Pintz in a paper “The difference between consecutive primes II” proved that for all $x > x_0$ the interval $[x-x^{0.525},x]$ contains prime numbers (with enough effort, the value of $x_0$ could be determined effectively). Thus $q=n^3-O(n^{1.575})$. Also, if $n\ge 2$ then by Bertrand’s postulate, $q\ge\frac {n^3}2$. Since all fractions $\frac rq$ with $1\le r<q$ are reduced and a difference between any two consecutive of them is $\frac 1q$, for each $k_n$ there exists $r$ such that $\left|x_{n,k_n}-\frac rq\right|\le \frac 1{2q}\le \frac 1{n^{3}}$.