# Prove the inequality - 1\4(ln2)^2 <= sigma(2^n\(2^(2^n)))

1. Aug 30, 2011

### puzzek

1. The problem statement, all variables and given/known data
This is a question taken from an old exam so I am not sure to which subject in calculus it's connected to...

Prove the inequality:
$\frac{1}{4(ln2)^2}$$\leq$$\sum\frac{2^n}{2^(2^n)}$

(sigma is from 1 to +inf, and the Denominator on the right side is (2^(2^n))

2. Relevant equations

3. The attempt at a solution

Well, I solved it by calculating the first 3 elements of the sum showing their sum is larger than the left side, and that proves it. But, I don't think that's the way I suppose to solve it!
(I thought it somehow connected to derivative of the sum of the integrals of the function in the sigma.

Any help would be greatly appreciated.

2. Aug 30, 2011

### PAllen

There is nothing wrong with what you've done (noting that all terms of the series are positive). However, if it is that trivial, I would double check that you have the left hand side right. I don't see that a teacher would assign something that trivial.

3. Aug 30, 2011

### puzzek

PAllen,

I double checked it again and it's exactly as it is written on my paper.

By the way, this is not that trivial, because in our test you are not allowed to use a calculator hence it's hard to calculate the value of (ln2)^2. One can still use some knowledge to evaluate it, but during an exam it's not that easy.

I still think I got it wrong and that's not the "proper" way to solve it. Is there any other direction?

4. Aug 30, 2011

### PAllen

What is strange is how 'not close' the left side is. If you only use ln2 to 3 significant digits, and do computations guaranteed to be bounded above on the left and below on the right, you still trivially prove the inequality with 3 terms. I'll have to leave it so someone else to propose a non-computational proof. I don't see one at the moment.