# Infinite product

1. Jun 6, 2004

### jet

Does anyone know how to compute product
a1*a2*...
where ai=1-x^i, x in (0,1)

2. Jun 6, 2004

### Janitor

If I ever knew how to do that sort of thing, I guess I've forgotten. I did a quickie spreadsheet and found these approximate results:

x=.25, product=0.688537537

x=.333333, product=0.56012661

x=.5, product=0.288788095

x=.666666, product=0.069273004

x=.75, product=0.015549724

I eyeballed the result for x=.5 in particular, and cannot think of any neato type of number that it corresponds to.

3. Jun 6, 2004

### Janitor

Also worth mentioning-

I once came across something called "the tower of exponents." Take a number and raise it to a power equal to the number. Now instead, take a number and raise it to a power equal to that number raised to the power of itself. Continue this indefinitely. The tower converges over some range of arguments. I can't remember the range, but its endpoints were based on Euler's number. Something along the lines of 1/e to 1/(e-1), that sort of thing.

4. Jun 6, 2004

### Zurtex

I've never ever done this so I am probably wrong, but is this what you want to know:

$$(1-x)(1-x^2)(1-x^3)...$$

Where $0 < x < 1$. Well correct me if I am wrong but does that not mean where $n \geq 1$, $|1 - x^n| < 1$. And thus an infinite product would equal 0?

Last edited: Jun 6, 2004
5. Jun 6, 2004

### matt grime

the product exists and is non-zero exactly when the sum of the a_is exists and none of the individual terms 1-a_i is zero. at least that is the theory of products.

in general the product is the limit of

$$1-\sum_{i=1}^{n}a_1 + \sum_{i<j\ i,j=1}^n a_ia_j + \ldots$$

as n tends to infinity

i think you might be able to work it out int this case though i don't know what the answer is.

6. Jun 6, 2004

### Hurkyl

Staff Emeritus
Zurtex, consider this product:

3/4 * 8/9 * 15/16 * 24/25 * ...

Note that the partial products are:

3/4
2/3
5/8
3/5
7/12
...

these partial products approach 1/2, despite each term being less than 1.

7. Jun 6, 2004

### Zurtex

Oh right thanks

8. Jun 6, 2004

### TALewis

I've never formally done infinite products, but isn't the following true:

$$\lim_{i\rightarrow\infty} 1-x^i = 1\, ,\quad 0<x<1$$

Eventually you're just multiplying by 1, so the infinite product should converge somewhere. Is that line of reasoning valid?

9. Jun 6, 2004

### Janitor

When I first read the thread-starter post, my worry was that all such products might converge trivially to zero. But then my spreadsheet and Hurkyl's post convinced me otherwise.

10. Jun 6, 2004

### Muzza

TALewis, I don't think you can split up the limit of a product that way (assuming I'm reading your mind correctly). Consider $$\lim_{x\rightarrow\infty} \frac{1}{x}x$$, it's not equal to $$\lim_{x\rightarrow\infty} \frac{1}{x} * \lim_{x\rightarrow\infty} x$$, is it?

Last edited: Jun 6, 2004
11. Jun 6, 2004

### TALewis

Well, I'm not saying this:

$$\lim_{i\rightarrow\infty} 1-x^i = \lim_{n\rightarrow\infty}\prod_{i=1}^n1-x^i$$

That's clearly untrue. I think I was trying to say that a_i approaches 1, so that a1*a2*a3... isn't going to be zero. But that could be an invalid leap in reasoning; I'm not sure. This isn't my strongest area.

12. Jun 6, 2004

### Muzza

That's not what I thought you were saying either. ;) But you can probably ignore my post anyway, I confused myself by using x (which was a constant in the original post). It would seem as if your thinking is correct, since the factors converges to 1, the product should converge too.

13. Jun 6, 2004

### matt grime

it is no more true, in fact is exactly as false as saying that an infinite sum converges if its terms got to zero as to say that a product converges if its terms go to 1. go back and read my post about 5 previously where I explained that.

14. Jun 6, 2004

### TALewis

I see. I guess it's similar to the divergence test for infinite series:

$$\sum_{i=1}^\infty a_i \mathrel{\mbox{is divergent if}} \lim_{i\rightarrow\infty}a_i \neq 0$$

However, this test cannot be used to show the convergence of a series, even though it might appear to "make sense," just as my previous proposition sounded correct to me at first. I stand corrected.

Last edited: Jun 6, 2004
15. Jun 6, 2004

### Muzza

Sorry Matt, I must've missed your post. Do you have an example of a product which diverges, but where the factors go to 1? (Not that I doubt your word or anything ;)).

16. Jun 6, 2004

### matt grime

Firstly let's be clear than when we say does not converge we mean, in this case, that the 'limit' is zero. after all it is not possible to have non-zero things multiplying to give zero. that's the rule of thumb we adopt (of course it may also wander off to infinity).

since the rule i gave was that 1-a_r has convergent product iff a_r has convergent sum the first thing you should think is: do i know any sums that misbehave?

the sum diverges, yet the terms go to zoer, similarly even though the terms 1-a_r tend to 1, the product is the limit

1/2*2/3*3/4....*(n-1)/n = 1/n which tends to zero. and that is your canonical non-example.

if you want something that is more convincing consider a_r=-1/r, where the product does tend to infinity.

17. Jun 6, 2004

### TALewis

Muzza, here's a real simple one I looked up:

\begin{align*} \lim_{k\rightarrow\infty}1+\frac{1}{k} &= 1\mbox{, whereas}\\ \prod_{k=1}^\infty 1+\frac{1}{k} &= \infty \end{align*}