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 am trying to prove following inequality:


I tried Stirling approximation but I could not get anything further. Then I get $$\binom{n}{k}\approx \frac{\sqrt{2\pi n}n^n}{2\pi \sqrt{k(n-k)}(n-k)^{n-k}k^k}$$

share|cite|improve this question
What have you tried? Please show your work so far. When you put in Stirling, you get some cancellation and a well known expression that goes to $e$. – Ross Millikan Jun 2 '11 at 14:36
You may even be able to get the much sharper $\tbinom{n}{k}<e(n/k)^k$ – Ross Millikan Jun 2 '11 at 14:49
@Ross: For $n=10$ and $k=3$, ${n \choose k}=120$ but $e(n/k)^k < 101$. – Shai Covo Jun 2 '11 at 15:32

$$\binom{n}{k} \left( \frac{k}{en} \right)^k = \frac{n(n-1) \ldots (n-k+1)}{n^k} \frac{k^k}{k! e^k} \leq \frac{k^k}{k! e^k} \text{ and since } e^k = \sum_m \frac{k^m}{m!},\;\;\; \frac{k^k}{k! e^k} < 1.$$

share|cite|improve this answer
I just edited your answer to enlarge it (it was so tiny!)...didn't change any notation or anything. I hope you don't mind? – amWhy Jun 2 '11 at 21:34
No problem, thanks! – Plop Jun 2 '11 at 21:40

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.