Prove that the limit of this sequence is > 0

[seeker101]

Any suggestions on how to prove that

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

 

[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 - 1/2^j only gets closer to 1 > 0.

 

[disregardthat]

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 - 1/2^j only gets closer to 1 > 0.

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.

 

[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

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

 

[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.

 

[disregardthat]

every term is positive so I can't imagine why the product wouldn't be positive also.

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.

 

[CRGreathouse]

every term is positive so I can't imagine why the product wouldn't be positive also.

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).

 

[fourier jr]

argh maybe I shouldn't have even replied

 

[CRGreathouse]

argh maybe I shouldn't have even replied

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

 

[seeker101]

Many thanks!

 

[hamster143]

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