# Prove that the limit of this sequence is > 0

1. Aug 16, 2010

### seeker101

Any suggestions on how to prove that

$$\prod^{\infty}_{j = 1} \left(1-\frac{1}{2^j}\right)$$ is greater than zero?

2. Aug 16, 2010

### CompuChip

How about proving that every term in the product is greater than zero?

I mean, the first one is 1 - 1/2 = 1/2 > 0, and as j gets larger, 1 - 1/2j only gets closer to 1 > 0.

3. Aug 16, 2010

### disregardthat

That's not sufficient. Suppose the term was n/(n+1). As n gets larger, this term gets closer to 1 > 0. And all terms are positive. That does not mean the infinite product $$\prod^{\infty}_{n=1} \frac{n}{n+1}$$ is positive. It is in fact 0.

4. Aug 16, 2010

### disregardthat

The product converges and has positive limit if and only if $$\sum^{\infty}_{k=1} -\log(1-2^{-k})$$ converges by taking the logarithm.

Using the taylor sum for the logarithm, the product converges if and only if
$$\sum^{\infty}_{k=1} \sum^{\infty}_{n=1} \frac{(2^{-k})^n}{n}$$
converges and is equal to it if so. Since all terms are positive, we can interchange the limits (using some result of analysis). That is, the previous expression converges if and only if
$$\sum^{\infty}_{n=1} \sum^{\infty}_{k=1} \frac{(2^{-k})^n}{n} = \sum^{\infty}_{n=1} \frac{\frac{1}{1-2^{-n}}-1}{n} = \sum^{\infty}_{n=1} \frac{1}{(2^n-1)n}$$
converges and is equal to it if it does. The equality above follows from the formula of the geometric series.

For n > 1, we have $$(2^n-1)n \geq 2^n+2^n-n \geq 2^n$$. The last expression converges if and only if $$\sum^{\infty}_{n=2} \frac{1}{(2^n-1)n} \leq \sum^{\infty}_{n=2} \frac{1}{2^n}$$ does, but this is equal to 1/2. So the other sum is less that $$\frac{1}{2}+\frac{1}{(2^1-1)1 } = \frac{3}{2}$$. This implies that

$$\sum^{\infty}_{k=1} -\log(1-2^{-k}) \leq \frac{3}{2} \Leftrightarrow \sum^{\infty}_{k=1} \log(1-2^{-k}) \geq -\frac{3}{2} \Leftrightarrow \prod^{\infty}_{k=1} (1-2^{-k}) \geq e^{-\frac{3}{2}}>0$$

Last edited: Aug 16, 2010
5. Aug 16, 2010

### fourier jr

isn't there a theorem that says $\prod_{n=1}^{\infty}(1+|a_n|)$ converges if & only if $\sum_{n=1}^{\infty}|a_n|$ converges? I think that would help somehow if a - sign were put somewhere or whatever. every term is positive so I can't imagine why the product wouldn't be positive also.

6. Aug 16, 2010

### disregardthat

Look at my counter-example for this line of arguing two posts up. That product behaves in a similar fashion, but can easily be seen to converge to 0.

7. Aug 16, 2010

### CRGreathouse

The equivalent thing for a sum would be "every summand is finite so I can't imagine why the sum wouldn't be finite also". Of course both are false:
$$\sum_{n=1}^\infty-1$$
diverges (to $-\infty$), and so does
$$\exp\left(\sum_{n=1}^\infty-1\right)=\prod_{n=1}^\infty e^{-1}$$
(to 0).

8. Aug 16, 2010

### fourier jr

argh maybe I shouldn't have even replied

9. Aug 16, 2010

### CRGreathouse

We're just trying to help with some (hopefully?) intuitive explanations.

10. Aug 18, 2010

### seeker101

Many thanks!

11. Aug 18, 2010

### hamster143

Just notice that $e^{-2x} < 1-x < e^{-x}$ for $0<x\leq 0.5$.