# Factorial Inequality problem $\left(\frac n2\right)^n > n! > \left(\frac n3\right)^n$

I met an inequality, I ask, do not mathematical induction to prove that:

Prove $\left(\frac n2\right)^n > n! > \left(\frac n3\right)^n$ without using induction

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Note that this does not work always, particularly when $n \in \{0,1,2,3,4,5\}$ – Aritra Das May 1 at 15:45

Let $a_n =\displaystyle \frac{2^n n!}{n^n}.$ Note that $a_6 = 80/81 < 1.$ We also have $$\frac{a_{n+1}}{a_n} = \frac{2^{n+1} (n+1)! }{(n+1)^{n+1}} \cdot \frac{n^n}{2^n n!} = 2 \left(\frac{n}{n+1}\right)^n < 1.$$ The sequence $$x_n = \left(1- \frac{1}{n+1} \right)^n$$ is monotonically decreasing to $1/e.$ Since $e>2$, $a_{n+1}/a_n < 1$ so $(a_n)$ is a monotonically decreasing sequence. Thus the first inequality holds.
By considering Taylor series, $\displaystyle e^x \geq \frac{x^n}{n!}$ for all $x\geq 0,$ and $n\in \mathbb{N}.$ In particular, for $x=n$ this yields $$n! \geq \left( \frac{n}{e} \right)^n$$ and this is stronger than the second inequality.
We could have used the same proof method for the second inequality as we did for the first: Let $b_n= \displaystyle \frac{3^n n!}{n^n}.$ Then $b_6 = 45/4 > 1.$ Also, $$\frac{b_{n+1}}{b_n} = \frac{3^{n+1} (n+1)! }{(n+1)^{n+1}} \cdot \frac{n^n}{3^n n!} = 3 \left(\frac{n}{n+1}\right)^n$$ and this is greater than $1$ since $e<3.$
What we have just done suggests we can prove the following: If $a,b$ are positive real numbers such that $a<e<b$ then $$\left(\frac{n}{a} \right)^n > n! > \left( \frac{n}{b} \right)^n$$ holds for sufficiently large $n$. This is turn suggests something stronger about the growth of the factorial function: $n!$ is asymptotically equal to $(n/e)^n$ to within at most a sub-exponential factor.
Typo : "$\frac{b_{n+1}}{b_n} = \ldots$ and this is greater than 1 since $e < 3$." – TenaliRaman May 12 '12 at 12:41