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

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: $$ \left( 1 + \frac{1}{n+1} \right)^{n+1} = \sum_{k=0}^{n+1} \binom{n+1}{k} \frac{1}{(n+1)^k} \ge \sum_{k=0}^{1} \binom{n+1}{k} \frac{1}{(n+1)^k} = 1 + (n+1) \frac{1}{n+1} = 2 $$


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

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


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

Tags:

Inequality

Factorial

Induction

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