Inequality $k!\pi+\frac{\pi}{6}\le{m!}\le{k!}\pi+\frac{5\pi}{6}$

Question

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Is the following statement is true?

$\forall k \in \mathbf{N},\exists{m}\in\mathbf{N}, k!\pi+\frac{\pi}{6}\le{m!}\le{k!\pi}+\frac{5\pi}{6}$

I tried by descendant proof but was not satisfied by the arguement.

Answer 1

If such $m$ exists, $k<m$ obviously, so $(k+1)!\le m!$. But if $k$ is sufficiently large, $(k+1)!>k!\pi+\frac56\pi$, which contradicts with the condition. Thus, the statement is false.

Answer 2

It's not true. There is no $m\in\mathbb N$ such that $$1!\cdot \pi+\frac{\pi}{6}\le m!\le 1! \cdot \pi+\frac{5}{6}\pi.$$ Here, note that $$1!\cdot \pi+\frac{\pi}{6}=\frac{7}{6}\pi\approx 3.7,\ \ 1!\cdot\pi+\frac{5}{6}\pi=\frac{11}{6}\pi\approx 5.8.$$