# Generalization of infinite product problem

1. Aug 20, 2010

### Fraqtive42

Prove that $$\prod_{j=1}^{\infty}\left(1-\frac{1}{f(j)}\right)>0$$ for all $$f:\mathbb{N}^{+}\to\mathbb{R}^{+}$$ which satisfy $$f(1)>1$$ and $$f(m+n)>f(m)$$, where $$m,n\in\mathbb{N}^{+}$$.

I found the problem "Prove that $$\prod_{j=1}^{\infty}\left(1-\frac{1}{2^{j}}\right)>0$$" and felt the need to make a generalization of it. So, here it is!

2. Aug 20, 2010

### g_edgar

Did you try f(n)=n+1 ??

3. Aug 20, 2010

### Fraqtive42

You are saying that f(n)=n+1 doesn't work?

4. Aug 20, 2010

### Fraqtive42

Hmm... that is weird. My proof had some kind of error in it...

5. Aug 21, 2010

### g_edgar

In fact the infinite product (1-1/(n+1)) diverges to zero.

6. Aug 21, 2010

### HallsofIvy

?? Do you mean converges to 0?

7. Aug 21, 2010

### Hurkyl

Staff Emeritus
I'm pretty sure it's standard to use "divergent" to describe an infinite product whose sequence of partial products converges to zero. It's directly analogous to the similar usage regarding infinite sums whose sequence of partial sums converges to $-\infty$.

8. Aug 22, 2010

### Fraqtive42

I think that I have made my proof on the assumption that for any functions satisfying these conditions, $$\prod_{j=1}^{N}(f(j)-1)\leq\prod_{j=1}^{N}f(j)$$ for any $$N\in\mathbb{N}$$. I don't see why $$f(n)=n+1$$ is an exception though.

9. Aug 22, 2010

### aq1q

I have never heard or seen it being used like that. I thought the general definition of convergence implied that the limit approached a finite value. But of course, I do understand what you are saying and it makes sense.

Last edited: Aug 22, 2010
10. Aug 22, 2010

### aq1q

This isn't true. As n gets larger, the partial products approach 1.

Ya, don't make that assumption. Just assume what's given: f(m+n)>f(m). Use induction to prove the original claim. Shouldn't be too difficult.

Last edited: Aug 22, 2010
11. Aug 22, 2010

### CRGreathouse

$$\prod_{n=1}^N\left(1-\frac{1}{n+1}\right)=\prod_{n=1}^N\frac{n}{n+1}=\frac12\frac23\frac34\cdots\frac{N}{N+1}=\frac{1}{N+1}$$

12. Aug 22, 2010

### aq1q

ah so sorry. I was accidentally taking the SUM. In that case, it DOES converges to 0. Perhaps, F needs to be clearly defined.

Last edited: Aug 23, 2010