Let $r\in [0,1]\setminus\mathbb{Q}$, let $\mathbb{N}$ denote the set of non-negative integers and let $\mathbb{N}\_+=\mathbb{N}\setminus\{0\}$. For $n\in\mathbb{N}\_+$ let $$\alpha\_r(n)=\min\{\big|r-\frac{m}{n}\big|: m\in\mathbb{N}\}$$ be the best approximation of $r$ that can be obtained using $n$ as the denominator.


For $k\in\mathbb{N}\_+$ we say that $r\in[0,1]\setminus\mathbb{Q}$ is *$k$-good* if there is $c\in\mathbb{N}\_+$ as well as a "starting index" $N\_0\in \mathbb{N}\_+$ such that $$\alpha\_r(n) < c \cdot \frac{1}{n^k} \;\text{ for all } n\geq N\_0.$$


Note that for all irrational $r$ in the unit interval, we have $\alpha\_r(n) < \frac{1}{n}$ for all $n\in\mathbb{N}\_+$, so every such $r$ is $1$-good.


For which integers $k \geq 2$ is there a $k$-good irrational number $r\in[0,1]\setminus\mathbb{Q}$? Is there even such an $r$ that is simultaneously $k$-good for all integers $k\geq 2$? (Only the first question needs to be answered for acceptance.)



Measuring the quality of real approximation
This certainly can't happen for any $k > 1$. (There is no reason to assume that $k$ is an integer rather than real number). In fact, it can't even happen if you just restrict the $n$ to sufficiently large powers of two.


Suppose you know that for every sufficiently large integer $i$ there is an integer $m$ such that


$$\left|r - \frac{m\_i}{2^i}\right| < \frac{c}{2^{ik}}.$$


Compare this appoximation with the approximation for $n = 2^{i+1}$. We have


$$\left|r - \frac{m\_{i+1}}{2^{i+1}}\right| < \frac{c}{2^{(i+1)k}}$$


By the triangle inequality


$$\left| \frac{m\_i}{2^i} - \frac{m\_{i+1}}{2^{i+1}} \right| < \frac{c}{2^{ik}} + \frac{c}{2^{(i+1)k}}.$$


If $k > 1$, then for large enough $i$ (depending only on $c$) the RHS will be strictly less than $1/2^{i+1}$. On the other hand the left hand side is equal to


$$\left| \frac{2m\_{i} - m\_{i+1}}{2^{i+1}} \right|$$


and this is at least $1/2^{i+1}$ if it is non-zero. Hence the only way the inequalities can hold for all sufficiently large $i$ is if eventually we have $2 m\_i = m\_{i+1}$ for all $i \ge i\_0$. But then our sequence of rational approximations to $r$ are all equal to the (rational) number $m\_{i\_0}/2^{i\_0}$, which implies that $r$ is rational.


