Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
This is tangential to the question, but it's very useful to know that $(n/3)^n \leq n! \leq (n/2)^n$ for all sufficiently large $n$. This is an extremely rough version of Stirling's formula, and in many applications it is all one needs. The inequalities can be derived by taking $k=2,3$ in applying the ratio test to the series $\sum_n (n/k)^n/n!$ (recall that $e = \lim_n (1 + 1/n)^n$). The same argument shows that $n! \geq (n/k)^n$ eventually holds if $k > e$, and the reverse eventually holds if $0 < k < e$. Seeing how $n!$ compares to $(n/e)^n$ is of course Stirling's formula territory. – leslie townes Oct 26 '11 at 22:13
up vote 7 down vote accepted

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

share|cite|improve this answer
Very cool, thanks a lot for breaking it down so much! – Martin Ueding Oct 26 '11 at 19:12


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

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

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

share|cite|improve this answer
Maybe I do not look right, but isn't that what I have above just in reverse? – Martin Ueding Oct 26 '11 at 19:00
I did just rewrite what you have to a certain extent. That's fair. But now it's clear exactly what you have to to fill in the gaps. – JavaMan Oct 26 '11 at 19:01
I was just too fast, now it helps. Thanks! – Martin Ueding Oct 26 '11 at 19:12

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

share|cite|improve this answer

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

share|cite|improve this answer

Your Answer


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.