Is it possible to turn infinite sums into infinite products?

Question

I am working on studying infinite products. I know that it is possible to convert an infinite product to an infinite sum using logarithms, where

$$\log \prod s_n = \sum \log s_n$$

My question is, it always possible go from sum to product with $e$, where

$$\exp \sum s_n = \prod e^{s_n}$$

Also, are there any other methods to convert sums into products?

Thanks in advance.

Answer 1

Yes it is possible. This is because $f = \exp(x)$ is a continuous homomorphism from the reals under addition to the reals under multiplication. That is, it is continuous and satisfies the identity $$f(x + y) = f(x)f(y)$$

Now for any such function we have that by definition of a infinite sum

$$ f\left( \sum_{n=1}^\infty s_n \right) = f\left(\lim_{N \to \infty} \sum_{n=1}^N s_n \right)$$

But now by continuity:

$$ f\left(\lim_{N \to \infty} \sum_{n=1}^N s_n \right) = \lim_{N \to \infty} f\left(\sum_{n=1}^N s_n \right) $$

And a finite application of the homomorphism property yields:

$$ \lim_{N \to \infty} f\left(\sum_{n=1}^N s_n \right) = \lim_{N\to\infty} \prod_{n=1}^N f(s_n)$$

But this is the definition of an infinite product. $$ \prod_{n=1}^\infty f(s_n)$$

Now, it turns out the only maps that satisfy these conditions (continuity and homomorpism) are of the form $f(x) = a^x$, for some positive $a$, but these are basically the same.

Answer 2

Under certain circumstances it is possible to factor a power series and write it as a infinite product of binomials.

An example is,

$$(1-x^2/\pi^2)(1-x^2/4\pi^2) \cdots = \sin(x)/x = 1 - \frac{1}{6} x^2 + \cdots, $$

this is kind of like the fundamental theorem of algebra for infinite polynomials.