Does the inequality $ n! > A \cdot B^{2n+1}$ hold for sufficiently large $n$?

Question

Suppose $A,B >0$ are given constants. Is it possible to find a large enough $n \in \mathbb{N}$ such that

$$ n! > A \cdot B^{2n+1}?$$

Answer 1

Pick $k$ large enough so that $2k+1>AB^3$ and $k>B^4$ and set $n=2k+1$. Then $$ n!>(2k+1)\times[(2k)\cdot(2k-1)\cdots(k+2)\cdot (k+1)]\\ >AB^3\times(B^4)^k=AB^3\times B^{4k}=AB^{4k+3}=AB^{2n+1}. $$

Answer 2

Yes. By Sterling's approximation, for large n, the factorial is approximately

$\sqrt{2\pi}*n^n/e^n$

Since A and B can be anything, it is valid to replace 2n+1 with n. The inequality becomes, for large n,

$\sqrt{2\pi}*n^n/e^n>A*B^n$

or

$n^n>(A/\sqrt{2\pi})*(eB)^n$

Which is clearly true for sufficiently large n.

If you want a proof without Sterling's approximation, simply consider the inequality for any given n. When going from n to n+1, the left side is multiplied by n+1 and the right side by $B^2$. Clearly, for sufficiently large n, $n+1>B^2$. Thus, once you reach this point the left side will start growing much faster than the right and will eventually catch up and, once this happens, the inequality will hold for all larger n.

Answer 3

Yes. For some $k$: $A<B^{k-1}$, So we'll be done if we can show that $n!>B^{2n+k}$.

If we select $n > 2B^2$, half the factors of $n!$ are larger than $B^2$ so the multiplum of those factors are larger than $B^{2n}$. It's not hard to correct for the last $k$ factors of $B$.

Answer 4

Let $C =\max \{ A,B \}$. If we prove $n! >C^{2n+2}$ for $ n >N_0$ we are done.

If $C< 1$ there is nothing to prove. Otherwise, see that for $n > 2C^6$ you have $$n!\geq 1 \cdot 2 \cdot .. \cdot \lfloor \frac{n}{2} \rfloor C^6 \cdot...C^6 \geq (C^6)^{\frac{n}{2}-1} $$

Now, it is easy to see that for $n >8$ we have $$(C^6)^{\frac{n}{2}-1} \geq C^{2n+2}$$

P.S. The proof uses the obvious inequality $n! \geq (\lfloor \frac{n}2 \rfloor)^{\frac{n}{2}-1}$, the rest is simple.