Generalization of infinite product problem

[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!

 

[g_edgar]

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

 

[Fraqtive42]

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

 

[Fraqtive42]

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

 

[g_edgar]

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

 

[HallsofIvy]

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

?? Do you mean converges to 0?

 

[Hurkyl]

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 $-\infty$.

 

[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.

 

[aq1q]

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 $-\infty$.

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.

 

[aq1q]

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

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

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.

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.

 

[CRGreathouse]

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

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

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

 

[aq1q]

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

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