RespeckKnuckles
New member
- Joined
- Sep 28, 2010
- Messages
- 2
I just can't figure this one out, been at it for hours. Need to prove:
\(\displaystyle \sum_{i=0}^{a} \binom{b}{i} \le (\frac{eb}{a})^a\)
With the two hints:
\(\displaystyle \sum_{i=0}^a \binom{b}{i} \le \sum_{i=0}^a (\frac{b}{a})^{(a-i)} \binom{b}{i}\)
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. Help!
\(\displaystyle \sum_{i=0}^{a} \binom{b}{i} \le (\frac{eb}{a})^a\)
With the two hints:
\(\displaystyle \sum_{i=0}^a \binom{b}{i} \le \sum_{i=0}^a (\frac{b}{a})^{(a-i)} \binom{b}{i}\)
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. Help!