how do you do infinite products?

Question

i don't know how to do infinite products. the three examples i have are:

Product of: (1-1/n^2)

Product of: ( (n^3-1)/(n^3+1) )

Product of: ( 1-x^(n^2) ) where |x|<1

for the first one i have a ridiculously over complicated method similar to the Wallis product but there must be another way...

Can anyone help?
Thanks

sorry that last one should be a +, not a - , and 2^n not n^2:
this is how it should look:

Product of: ( 1+x^(2^n) ) where |x|<1

Answer 1

1)
1-1/n^2 = (n-1)(n+1)/n^2

when you write the product, terms at denominators will simplify with terms at numerators of succesive fractions, resulting only a finite number of them . That is called telescopic product i guess

2) that is equal with

(n-1)(n^2+n+1)/((n+1)(n^2-n+1)) =

(n-1)/(n+1) * (n^2+n+1)/(n^2-n+1)
notice that (n+1)^2-(n+1)+1 = n^2+n+1.

so again, when you write the product, terms from denominators will simplify with terms from numerators

3) (1+x)(1+x^2) = 1+x+x^2+x^3
(1+x)(1+x^2)(1+x^4) = 1+x+x^2+x^3+x^4+x^5+x^6+x^7

what did you notice?
Prove for general case using induction.

Answer 2

Look at the first few partial products.

(I assume that for the first two, n starts with 2; and that for the last one, n starts with 0.)

The first product gives the partial products (after reducing to lowest terms)

3/4, 2/3, 5/8, 3/5, 7/12, 4/7, 9/16, ...

These appear to jump around, but consider the equivalent

3/4, 4/6, 5/8, 6/10, 7/12, 8/14, 9/16, ...

The numerators form an arithmetic sequence, and so do the denominators. Find linear expressions for the numerators and denominators, then use induction to prove that the formula holds for all n. Then the limit (which is 1/2) is easy to show.

And in fact, knowing the limit is 1/2, you will see that the partial products may be written

1/2 + 1/4, 1/2 + 1/6, 1/2 + 1/8, 1/2 + 1/10, etc.

making a formula for the nth partial product very easy to obtain.

Similarly for the second sequence of partial products:

7/9, 13/18, 7/10, 31/45, ...

but experiment a bit and you get the equivalent

7/9, 13/18, 21/30, 31/45, ...

This time the numerators form a quadratic sequence, and so do the denominators. Handle this product similarly to the first one. And again, using the limit of 2/3, note that the sequence of partial products may be written

2/3 + 1/9, 2/3 + 1/18, 2/3 + 1/30, 2/3 + 1/45, etc.

so finding a formula for the nth partial product is a bit easier.

I'm sure there must be a more methodical way to do this, but I don't know it.

For the third, multiply out the partial products; you'll get

1 + x, 1 + x + x² + x³, 1 + x + x² + x³ + x^4 + x^5 + x^6 + x^7, ...