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.

Show that $$n^{(l)}=n(n-1)\dots(n-l+1)\ge {n^l\over e}$$ where $2\le l \le \sqrt{n}$

Here is how far I've got with this:

$$n^{(l)}=\prod_{i=0}^{l-1}(n-i)=n^l\prod_{i=1}^{l-1}(1-{i\over n})\\ \ge n^l (1-{l-1\over n})^{l-1}\ge n^l (1-{l-1\over n})^{n} $$

Unfortunately this is not in the right form to apply a bound on the exponential function. Also I am worried that I haven't used the constraint on $l$ either.

Could anyone help me with this?

share|improve this question

1 Answer 1

up vote 1 down vote accepted

By convexity, on $0\le x\le \frac12$, $1-x\ge e^{-2x}$. Then after your first line, observe that in the product, $i\le \sqrt{n}-1\le \frac n2$ and so $$ \prod_{1\le i\le l-1} (1-\frac in)\ge \prod_{1\le i\le l-1} e^{-2i/n} = \exp -\frac 2n \frac{l(l-1)}{2}\ge \exp -1. $$

share|improve this answer

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.