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.

I would like to show that:

$$ \prod_{i=0}^{n} {n \choose i}\leq \left(\frac{2^n-2}{n-1}\right)^{n-1} $$

$$n=2p $$ $$ \prod_{i=0}^{2p} {2p \choose i}=\prod_{i=1}^{2p-1} {2p \choose i}\leq {2p\choose p}^{2p-1} $$

$$ \left(\frac{4^p-2}{2p-1}\right)^{2p-1}=\left(\frac{1}{2p-1}\sum_{k=1}^{2p-1}2^k\right)^{2p-1} $$

It suffices to show that:

$$ {2p\choose p}\leq \frac{4^p-2}{2p-1} $$

$$ n=2p+1 $$

It suffices to show that:

$$ {2p+1\choose p}\leq \frac{4^p-1}{p}=\frac{1}{2p}\sum_{k=1}^{2n} 2^k $$


$$ \sqrt[n-1]{{n\choose 1}...{n\choose n-1}}\leq\frac{{n\choose 1}+...+{n\choose n-1}}{n-1}=\frac{2^n-2}{n-1} $$

share|improve this question
add comment

2 Answers 2

up vote 2 down vote accepted

Your approach will not work, because the central binomial coefficient satisfies $\binom{2p}{p} = \Theta \left( \frac{2^{2p}}{\sqrt{p}} \right)$, which is asymptotically larger than $\frac{2^{2p} - 2}{p-1}$.

HINT for the correct approach: Apply AM-GM inequality to the $n-1$ binomial coefficients $\binom{n}{i}$ where $1 \leqslant i \leqslant n-1$.

share|improve this answer
Thanks! Actually the inequality is obvious, sorry... –  Chon Dec 29 '11 at 14:19
add comment

Unless I made a mistake you won't be able to prove your result with this approach since your formula :

$$ {2p+1\choose p}\leq \frac{4^p-1}{p}=\frac{1}{2p}\sum_{k=1}^{2n} 2^k $$

is not true for p = 2

$$ {2*2+1\choose 2} = {5\choose 2} = 10 \geq \frac{4^2-1}{2} = 15/2 $$

share|improve this answer
add comment

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.