Proving an Inequality by Induction: $n! < (n/2)^n$

Question

Prove by mathematical induction that for $n>5$ the following holds:

$$n! < (n/2)^n$$

I have been struggling with this problem; I have got to the point where I can show that $(n+1)! < (n+1)^n+1$, however, I do not know how to proceed any further.

Answer 1

If $P(n)$ is the proposition you can show directly $P(6)$ is true.

Now, $P(k)$ is $k! < (k/2)^k$ which implies $k!(k+1) < (k/2)^k(k+1)$ and

$$ \begin{align} \left( \frac k2 \right)^k (k+1) &\leq \left( \frac{k+1}{2} \right)^{k+1} \ \Leftrightarrow \\k + 1 &\leq \frac{k+1}{2} \left( 1 + \frac 1k \right)^k \Leftrightarrow \\ 2 &\leq \left( 1 + \frac 1k \right)^k \end{align} $$

It is well known that $a_k = \left( 1 + \frac 1k \right)^k$ is a monotonically increasing sequence converging to $e$, greater than $2$ for all $k \geq 2$. Hence $P(k) \Rightarrow P(k+1)$ for all $k \geq 6$.

Therefore $P(n)$ for all $n \geq 6$.

Answer 2

$$6!=720<729=3^6$$ Assume $n!<(\frac{n}{2})^n$ $$(n+1)!=(n+1)\cdot n! < (n+1) (\frac{n}{2})^n < \frac{n+1}{2} (\frac{n+1}{2})^n=(\frac{n+1}{2})^{n+1}$$ hte last inequality ocures because $$(n+1)^n=\sum_{i=0}^n \binom{n}{i}n^i$$ which gives a lot of extra terms that we don't care for here but that they add a lot extra and then dividing our intial $n+1$ by $2$ doesn't change that as the rest adds sufficiently much.

Answer 3

To prove $n!<(\frac{n}{2})^n$ is the same as to prove $2^n n!<n^n$.

1) For $n=6$ this is true.

2) Suppose that $2^n n!<n^n$ is true for other $n$ beyond $n=6$.

3) We must prove that if it is true for $n$ that then it implies that it works also for $n+1$.

Proof: $2^{n+1}(n+1)!=2(n+1)2^nn!<2(n+1)n^n<(n+1)(n+1)^n=(n+1)^{n+1}$.

Maybe the only non-obvious step in this chain of inequalities is $2n^n<(n+1)^n$ which is the same as $2n^n<\sum_{i=0}^n \binom {n}{i} {n}^{n-i}$ and this is true if and only if $n^n<\sum_{i=1}^n \binom {n}{i} {n}^{n-i}$ and this is true if and only if $0<\sum_{i=2}^n \binom {n}{i} {n}^{n-i}$ and the last one inequality is obviously true because on the right side are strictly positive terms.