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

there

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.

• How can it be true ? $\frac{m!}{k!}$, a product of integers, must be very close to $\pi$ ?! – Yves Daoust Aug 8 '14 at 9:32
• Did you try picking some $k$, say, $k=4$, and seeing whether there's an $m$? – Gerry Myerson Aug 8 '14 at 9:48
• every irrational can be arbitarary approximate by rationals. so may be it can be happan. – Bumblebee Aug 8 '14 at 9:58
• @Nilan, get back to me after you have found an integer $m$ such that $$24\pi+(\pi/6)\le m!\le24\pi+(5\pi/6)$$ – Gerry Myerson Aug 8 '14 at 23:33

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.
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 $$1!\cdot \pi+\frac{\pi}{6}=\frac{7}{6}\pi\approx 3.7,\ \ 1!\cdot\pi+\frac{5}{6}\pi=\frac{11}{6}\pi\approx 5.8.$$