Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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

share|improve this question
7  
What have you tried? Your questions always read: "how to prove (fill in something homeworkish)." Please show a little bit of your own efforts and read meta.math.stackexchange.com/q/1803/5363 –  t.b. Aug 13 '12 at 4:55
    
Are you allowed to use Stirling? –  J. M. Aug 13 '12 at 4:58
    
@J.M. This follows from AM GM –  Pedro Tamaroff Aug 13 '12 at 4:58
    
Stirling here seems like a pretty big hammer to squash a little bug. –  MJD Aug 13 '12 at 4:58
    
@Peter, I thought of that a few seconds after writing that comment. I'm getting lazy... –  J. M. Aug 13 '12 at 5:02
show 2 more comments

3 Answers

up vote 8 down vote accepted

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.

share|improve this answer
add comment

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.

share|improve this answer
    
@Peter: Yes, though I’m in the middle of writing up a couple of answers. –  Brian M. Scott Aug 13 '12 at 5:24
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.