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If $n>1$ is a natural number, how to prove that

$(n+1)^n > 2^n n!$

3 Answers 3

11

By the AM GM inequality, $\eqalign{ & \root n \of {n!} = \root n \of {1 \cdot 2 \cdot 3 \cdot 4 \cdots n} \leq \frac{{1 + 2 + 3 + \cdots + n}}{n} \cr & \frac{{\left( {n + 1} \right)n}}{{2n}} = \frac{{n + 1}}{2} \cr} $

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

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Prove it by induction on $n$. The base case $n=2$ is trivial. For the induction step you want to show that if $(n+1)^n>2^nn!$, then $(n+2)^{n+1}>2^{n+1}(n+1)!$. This will certainly be true if

$\frac{(n+2)^{n+1}}{(n+1)^n}\ge \frac{2^{n+1}(n+1)!}{2^nn!}=2(n+1)\;.\tag{1}$

The inequality $(1)$ is equivalent to the inequality

$\left(\frac{n+2}{n+1}\right)^{n+1}=\left(1+\frac1{n+1}\right)^{n+1}\ge 2\;.$

Now use the binomial theorem.

  • 0
    @Peter: Yes, though I’m in the middle of writing up a couple of answers.2012-08-13
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Let's consider the following auxiliary sequence: $u_{n}=\frac{(n+1)^n}{2^n n!}$ Then we notice the sequence is strictly increasing $ \frac{u_{n+1}}{u_{n}}=\frac{(n+2)^{n+1}}{2^{n+1} (n+1)!} \cdot \frac{2^n n!}{(n+1)^n}= \frac{1}{2} \left(1+\frac{1}{n+1}\right)^{n+1} > 1\space \ \tag1 $ and $u_{2}= \frac{9}{8} \tag2$

From $(1)$ and $(2)$ we get the needed inequality $(n+1)^n > 2^n n!$

Q.E.D.