What is the corresponding infinite series for this infinite infinite product?

Solution 1:

The answer obviously is:

$$1+{x\over\alpha-1}+{x^2\over(\alpha-1)\cdot(\alpha^2-1)}+\dots+{x^n\over(\alpha-1)\cdot...\cdot(\alpha^n-1)}+\dots$$

How so?

Well, what is the coefficient at $x^n$? That's an infinite sum of various inverse powers of $\alpha$, some of them identical. How many terms $\frac1{\alpha^m}$ are there? As many as there are sets of $n$ different integers that sum to $m$. Is there any shorter way to express this? Why, sure; it is the same as in the problem "in how many ways can you give such-and-such sum in change, if you have such-and-such coins".

Taking the term at $x^2$ as an example:

$${1\over(\alpha-1)(\alpha^2-1)} = {1\over\alpha^3}\cdot\frac1{\left(1-{1\over\alpha}\right)\left(1-{1\over\alpha^2}\right)} = \\= \left({1\over\alpha}+{1\over\alpha^2}+{1\over\alpha^3}+\dots\right)\left({1\over\alpha^2}+{1\over\alpha^4}+{1\over\alpha^6}+\dots\right)= \\ = {1\over\alpha^3}+{1\over\alpha^4}+{2\over\alpha^5}+{2\over\alpha^6}+{3\over\alpha^7}+{3\over\alpha^8}+\dots$$ which is the same as I said earlier: the sum of inverse powers of $\alpha$ with certain combinatorial expression in the numerator.

So it goes.