Hi there,

Could anyone give me some tips on how to show for

I have no idea where to start. I expanded it out and you can see that the 1s cancel, but past that I am lost...

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- Aug 23rd 2010, 10:06 AMgaloisA bound on a product
Hi there,

Could anyone give me some tips on how to show for

I have no idea where to start. I expanded it out and you can see that the 1s cancel, but past that I am lost... - Aug 24th 2010, 02:16 AMred_dog
We have to prove the inequality

, for all integers k such as

We use the mathematical induction.

, which is true.

Suppose is true and we prove that is also true.

We have:

- Aug 24th 2010, 01:39 PMBruno J.
Here's another idea. For each , let denote a random variable which takes the value with probability , and the value otherwise. Suppose moreover that these random variables are independent. Then is the probability that all of the take the value , hence is the probability that not all of the take on the value , i.e. that at least one of them take the value . This is clearly less than .