A question about an infinite product

[WACG]

I am puzzled by the following infinite product:
Let B > A

A - B = [A^(1/2) + B^(1/2)] * [A^(1/2) - B^(1/2)]
= [A^(1/2) + B^(1/2)] * [A^(1/4) + B^(1/4)] * [A^(1/4) - B^(1/4)]

=[A^(1/2) + B^(1/2)] * [A^(1/4) + B^(1/4)] * [A^(1/8) + B^(1/8)] * [A^(1/8) - B^(1/8)]

etc.

Continuing the obvious expansion into an infinite product produces a sequence of terms none of which are negative. However, since B > A then A - B is a negative value. How can a infinite product of terms greater than zero produce a negative value? Surely there is a "simple" explanation.

Thanks for any comments.

 

[ramsey2879]

I am puzzled by the following infinite product:
Let B > A

A - B = [A^(1/2) + B^(1/2)] * [A^(1/2) - B^(1/2)]
= [A^(1/2) + B^(1/2)] * [A^(1/4) + B^(1/4)] * [A^(1/4) - B^(1/4)]

=[A^(1/2) + B^(1/2)] * [A^(1/4) + B^(1/4)] * [A^(1/8) + B^(1/8)] * [A^(1/8) - B^(1/8)]

etc.

Continuing the obvious expansion into an infinite product produces a sequence of terms none of which are negative. However, since B > A then A - B is a negative value. How can a infinite product of terms greater than zero produce a negative value? Surely there is a "simple" explanation.

Thanks for any comments.

In each of the products the last term to the right is negative.

 

[willem2]

well there's that last term [A^(1/(2^n) - B^(1/(2^n)], you can't ignore it,
unless you prove that its limit as n->infinity is equal to 1, and it isn't, because it's always negative.

 

[dodo]

Hi, WACG,
you are proposing thatA-B = \lim_{n \to \infty} \prod_{i=1}^n (A^{1/2^i} + B^{1/2^i})when in fact this limit does not exist: the factor A^{1/2^i} + B^{1/2^i} tends to 1+1 = 2 on infinity.

Each of the successive expressions in your original post is made of a larger and larger number (the product of the positives) times a smaller and smaller negative number.

The guys at the Calculus & Analysis forum would probably have something better to say, though.

 

[WACG]

Well, I guess my question becomes - How DO you write the indicated infinite product that resolves itself to A-B? In other words, how do you write the product that has the persistent and ever smaller negative number at the tail end?

 

[dodo]

I suppose I'd say that, for all integers n > 0,A-B = \left( \prod_{i=1}^n (A^{1/2^i} + B^{1/2^i}) \right) \cdot (A^{1/2^n} - B^{1/2^n})which is pretty much what you wrote -- before trying to produce a limit. The statement is true for all given n, but no limit needs to be involved, as I see it.

For another example, all of the products 1.(1/1), 2.(1/2), 4.(1/4), 8.(1/8), ... equal 1; the fraction is getting smaller and smaller, but that doesn't mean that the sequence 1,2,4,8,... converges, let alone to 1.

 

[WACG]

So you are telling me that a valid expression for A - B cannot be written as an infinite product (regardless whether it needs to be "involved").

 

[dodo]

Well, if by an "infinite product" you mean a limit, then no, since the limit does not exist. Presently I can't see any other way of moving from a finite product to an infinite one than by taking a limit, but mind openers are welcome.

P.S.: Obviously, the expression for A-B in post#6 *has* a limit when n->inf, since it has a constant value regardless of n. But if you want to get rid of the difference at the end and use only the product of the positive sums, that one has no limit.

P.P.S: In yet other words, the limit of a product of sequences is the product of their limits, provided these limits exist; in this case, one of them does not.