$n! \leq \left( \frac{n+1}{2} \right)^n$ via induction

Question

I have to show $n! \leq \left( \frac{n+1}{2} \right)^n$ via induction.

This is where I am stuck:

$$\left( \frac{n+2}{2} \right)^{n+1} \geq \dots \geq =2 \left( \frac{n+1}{2} \right)^{n+1} = \left( \frac{n+1}{2} \right)^n(n+1) \geq n!(n+1) = (n+1)! $$

I approached this from both sides and this is the closest I can get. I realize that $n+2$ on the left has to be bigger than $n+1$ on the right, but I do not know who to show that it overpowers the factor two I have from the right.

What could I do to fill the dots? Currently, I just have it without the dots, but I would be happier if I could back it up.

Answer 1

Assuming $n! \le \left( \frac{n+1}{2} \right)^n$ is true, carry the induction step

$$ (n+1) n!\leq (n+1) \left(\frac{n+1}{2}\right)^n =2 \left(\frac{n+1}{2}\right)^{n+1} \stackrel{?}{\leq} \left(\frac{n+2}{2}\right)^{n+1} $$ But the last inequality is just $$ 2 \le \left( \frac{n+2}{n+1} \right)^{n+1} = \left( 1 + \frac{1}{n+1} \right)^{n+1} $$ It follows because: $$ \left( 1 + \frac{1}{n+1} \right)^{n+1} = \sum_{k=0}^{n+1} \binom{n+1}{k} \frac{1}{(n+1)^k} \ge \sum_{k=0}^{1} \binom{n+1}{k} \frac{1}{(n+1)^k} = 1 + (n+1) \frac{1}{n+1} = 2 $$

Answer 2

Hint:

$$ (n+1)! = (n+1) n! \leq (n+1) \left( \frac{n+1}{2} \right)^n = 2 \left( \frac{n+1}{2} \right)^{n+1}. $$

You can check that $2 \left( \frac{n+1}{2} \right)^{n+1} \leq \left( \frac{n+2}{2} \right)^{n+1}$, by proving that

$$ 2 \leq \left( \frac{n+2}{n+1} \right)^{n+1}. $$

Answer 3

Hint: $$\left(\frac{n+2}{2}\right)^{n+1}=\frac{n+2}{2}\left(\frac{n+2}{n+1}\right)^n\left(\frac{n+1}{2}\right)^n.$$

Estimate $\left(\frac{n+2}{n+1}\right)^n$.

Answer 4

$\frac{((n+2)/2)^{n+1}}{((n+1)/2)^{n+1}} = (1 + \frac{1}{n+1})^{n+1} \ge 2$ by the binomial theorem.