Originally Posted by

**RespeckKnuckles** I just can't figure this out. Can someone point me in the right direction? I need to prove, that for all b >= a:

$\displaystyle

\sum_{i=0}^{a} \binom{b}{i} \le (\frac{eb}{a})^a

$

Two hints I was given were (first one is possible because b > a):

$\displaystyle

\sum_{i=0}^a \binom{b}{i} \le \sum_{i=0}^a (\frac{b}{a})^{(a-i)} \binom{b}{i}

$

and

$\displaystyle

(1 + \frac{1}{x})^x \le e

$

I've tried everything I could think of...using the binomial formula, expanding the first hint, etc. Always seem to hit dead ends.