General integer roots $\sqrt[k]{n}$ approximating $\pi$

Question

Earlier today when I was playing around with one of my homebuilt numerical $k$:th-root-solvers, I found that the number $\sqrt[3]{31} = 3.14138\cdots \approx \pi$ . A quite good approximation to $\pi$, actually. I got curious if there are some results in general for how good approximations for $\pi$ one can get if one is limited to perform $$\sqrt[k]{n}, \hspace{1cm} k,n\in \mathbb N$$ Can we derive some bound in terms of $k$ and $n$ for how close we can get to $\pi$? Also, is it a coincidence that $n=2^m-1$ gives a good value for some $k,m$ or can we show that those $n$ are extra fruitful somehow? Anything is welcome, but I would guess that geometric arguments on a circle could be extra useful(?)

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EDIT an even better approximation (correct decimal digits underlined) $$(31^4+748)^{1/12}= \underline {3.1415926}0 \cdots$$ Which was found by exploring $$(31^k+n)^{1/(3k)}\cdots$$

However here the $n$ seems to in general get much larger, which is not so simple or elegant.

Answer 1

Aproximations can get as good as you want. In fact, it is easy to see that $$A=\{\sqrt[k]n \in \mathbb R\colon n,k\in \mathbb N\}$$ is dense in $[1,\infty)$.

If we want to have $$\alpha-\varepsilon<\sqrt[k]n<\alpha+\varepsilon$$ for $\alpha\ge1$ and $\varepsilon >0$, then we need $k$ and $n$ such that $$(\alpha-\varepsilon)^k<n<(\alpha+\varepsilon)^k.$$

Under our assumptions it can be proved that $$(\alpha+\varepsilon)^k-(\alpha-\varepsilon)^k \to \infty$$ as $k \to \infty$, so in particular this difference can be made greater than one by taking $k$ large enough.

Let $k_0$ be such a value of $k$, then we must have some $n_0\in \mathbb N$ such that $(\alpha-\varepsilon)^{k_0}<n_0<(\alpha+\varepsilon)^{k_0}$ (for instance, at least one of $\left \lfloor (\alpha+\varepsilon)^{k_0} \right \rfloor$ and $\left \lceil (\alpha-\varepsilon)^{k_0} \right \rceil$ must work as such an $n_0$).

Then we have $$\alpha-\varepsilon<\sqrt[k_0]{n_0}<\alpha+\varepsilon$$ and since $\alpha\ge 1$ and $\varepsilon >0$ were arbitrary, it turns out that $A$ is dense in $[1,\infty)$.